{"id":30503,"date":"2022-01-30T13:41:27","date_gmt":"2022-01-30T08:11:27","guid":{"rendered":"https:\/\/infinitylearn.com\/surge\/?p=30503"},"modified":"2025-10-23T14:20:08","modified_gmt":"2025-10-23T08:50:08","slug":"study-materials-maths-introduction-to-trigonometry-class-10-extra-questions-maths-chapter-8","status":"publish","type":"post","link":"https:\/\/infinitylearn.com\/surge\/study-materials\/maths\/introduction-to-trigonometry\/class-10-extra-questions\/maths-chapter-8\/","title":{"rendered":"Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_37 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" style=\"display: none;\"><label for=\"item\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input type=\"checkbox\" id=\"item\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1' style='display:block'><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/maths\/introduction-to-trigonometry\/class-10-extra-questions\/maths-chapter-8\/#NCERT_Class_10_Maths_Chapter_8_-_Introduction_to_Trigonometry\" title=\"NCERT Class 10 Maths Chapter 8 \u2013 Introduction to Trigonometry\">NCERT Class 10 Maths Chapter 8 \u2013 Introduction to Trigonometry<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/maths\/introduction-to-trigonometry\/class-10-extra-questions\/maths-chapter-8\/#FAQs_on_Introduction_to_Trigonometry_-_Class_10_Maths_Chapter_8\" title=\"FAQs on Introduction to Trigonometry &#8211; Class 10 Maths Chapter 8\">FAQs on Introduction to Trigonometry &#8211; Class 10 Maths Chapter 8<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/maths\/introduction-to-trigonometry\/class-10-extra-questions\/maths-chapter-8\/#What_is_Trigonometry_in_Class_10_Maths\" title=\"What is Trigonometry in Class 10 Maths?\">What is Trigonometry in Class 10 Maths?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/maths\/introduction-to-trigonometry\/class-10-extra-questions\/maths-chapter-8\/#Why_is_Trigonometry_important_in_Class_10_Maths\" title=\"Why is Trigonometry important in Class 10 Maths?\">Why is Trigonometry important in Class 10 Maths?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/maths\/introduction-to-trigonometry\/class-10-extra-questions\/maths-chapter-8\/#What_concepts_are_covered_in_Class_10_Trigonometry\" title=\"What concepts are covered in Class 10 Trigonometry?\">What concepts are covered in Class 10 Trigonometry?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/maths\/introduction-to-trigonometry\/class-10-extra-questions\/maths-chapter-8\/#What_are_Trigonometric_Ratios\" title=\"What are Trigonometric Ratios?\">What are Trigonometric Ratios?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/maths\/introduction-to-trigonometry\/class-10-extra-questions\/maths-chapter-8\/#How_can_Trigonometry_be_applied_in_real_life\" title=\"How can Trigonometry be applied in real life?\">How can Trigonometry be applied in real life?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/maths\/introduction-to-trigonometry\/class-10-extra-questions\/maths-chapter-8\/#What_is_Pythagoras_Theorem_in_Trigonometry\" title=\"What is Pythagoras Theorem in Trigonometry?\">What is Pythagoras Theorem in Trigonometry?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-9\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/maths\/introduction-to-trigonometry\/class-10-extra-questions\/maths-chapter-8\/#How_are_Trigonometric_Ratios_calculated\" title=\"How are Trigonometric Ratios calculated?\">How are Trigonometric Ratios calculated?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-10\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/maths\/introduction-to-trigonometry\/class-10-extra-questions\/maths-chapter-8\/#What_are_the_primary_applications_of_Trigonometry\" title=\"What are the primary applications of Trigonometry?\">What are the primary applications of Trigonometry?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-11\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/maths\/introduction-to-trigonometry\/class-10-extra-questions\/maths-chapter-8\/#How_can_I_understand_Trigonometry_concepts_better\" title=\"How can I understand Trigonometry concepts better?\">How can I understand Trigonometry concepts better?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-12\" href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/maths\/introduction-to-trigonometry\/class-10-extra-questions\/maths-chapter-8\/#Is_Trigonometry_covered_in_other_math_classes_too\" title=\"Is Trigonometry covered in other math classes too?\">Is Trigonometry covered in other math classes too?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<p>Important questions of Chapter 8 \u2013 Introduction to Trigonometry of Class 10 are given here for students who want to score high marks in their board exams 2021-22. Students can also download the trigonometry class 10 questions which are available on our website. They can learn and solve the problems offline by downloading trigonometry class 10 questions.  The questions which are provided below are as per the latest CBSE syllabus and are designed according to the NCERT book. The questions are formulated after analyzing the previous year\u2019s questions papers, exam trends and latest sample papers. Solving these questions will help students to get prepared for the final exam. Students can also get important questions for all the chapters of 10th standard Maths. Solve them to get acquainted with the various types of questions to be asked from each chapter of the Maths subject.<\/p>\n<p>In Chapter 8, students will be introduced to the Trigonometry concept, which states the relationship between angles and sides of a triangle. They will come across many trigonometric formulas which will be used to solve numerical problems. The six primary trigonometric ratios are sine, cosine, tangent, secant, cosecant and cotangent. The whole trigonometry concept revolves around these ratios sometimes also called functions. Students can learn more about trigonometry class 10 questions, which are provided with complete explanations. <span style=\"color: #0000ff;\"><strong>Introduction to Trigonometry Class 10 Maths Chapter 8<\/strong><\/span> Important Questions are solved by our expert teachers so that students can understand the problems quickly. Besides, students can also get additional questions on chapter 8 of class 10 maths for practice at the end.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Class_10_Maths_Chapter_8_-_Introduction_to_Trigonometry\"><\/span>NCERT Class 10 Maths Chapter 8 \u2013 Introduction to Trigonometry<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>1. In \u2206 ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:<\/strong><\/p>\n<p><strong>(i) sin A, cos A<\/strong><\/p>\n<p><strong>(ii) sin C, cos C<\/strong><\/p>\n<p>Solution:<\/p>\n<p>In a given triangle ABC, right angled at B = \u2220B = 90\u00b0<\/p>\n<p>Given: AB = 24 cm and BC = 7 cm<\/p>\n<p>According to the Pythagoras Theorem,<\/p>\n<p>In a right- angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides.<\/p>\n<p>By applying Pythagoras theorem, we get<\/p>\n<p>AC<sup>2<\/sup>=AB<sup>2<\/sup>+BC<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup> = (24)<sup>2<\/sup>+7<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup> = (576+49)<\/p>\n<p>AC<sup>2<\/sup> = 625cm<sup>2<\/sup><\/p>\n<p>AC = \u221a625 = 25<\/p>\n<p>Therefore, AC = 25 cm<\/p>\n<p>(i) To find Sin (A), Cos (A)<\/p>\n<p>We know that sine (or) Sin function is the equal to the ratio of length of the opposite side to the hypotenuse side. So it becomes<\/p>\n<p>Sin (A) = Opposite side \/Hypotenuse = BC\/AC = 7\/25<\/p>\n<p>Cosine or Cos function is equal to the ratio of the length of the adjacent side to the hypotenuse side and it becomes,<\/p>\n<p>Cos (A) = Adjacent side\/Hypotenuse = AB\/AC = 24\/25<\/p>\n<p>(ii) To find Sin (C), Cos (C)<\/p>\n<p>Sin (C) = AB\/AC = 24\/25<\/p>\n<p>Cos (C) = BC\/AC = 7\/25<\/p>\n<p style=\"text-align: center;\"><strong>Also Check:<a href=\"https:\/\/infinitylearn.com\/surge\/study-material\/cbse-notes\/class-10\/maths\/important-questions-for-class-10-maths-chapter-14-statistics\/\" target=\"_blank\" rel=\"noopener\"> Important Questions for Class 10 Maths Chapter 14 Statistics<\/a><\/strong><\/p>\n<p><strong>2. In Fig. 8.13, find tan P \u2013 cot R<\/strong><\/p>\n<p><img loading=\"lazy\" class=\"alignnone size-full wp-image-711433\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2022\/01\/Screenshot-2024-03-07-123812.png\" alt=\"\" width=\"216\" height=\"211\" \/><\/p>\n<p>Solution:<\/p>\n<p>In the given triangle PQR, the given triangle is right angled at Q and the given measures are:<\/p>\n<p>PR = 13cm,<\/p>\n<p>PQ = 12cm<\/p>\n<p>Since the given triangle is right angled triangle, to find the side QR, apply the Pythagorean theorem<\/p>\n<p>According to Pythagorean theorem,<\/p>\n<p>In a right- angled triangle, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides.<\/p>\n<p>PR<sup>2<\/sup> = QR<sup>2<\/sup> + PQ<sup>2<\/sup><\/p>\n<p>Substitute the values of PR and PQ<\/p>\n<p>13<sup>2 <\/sup>= QR<sup>2<\/sup>+12<sup>2<\/sup><\/p>\n<p>169 = QR<sup>2<\/sup>+144<\/p>\n<p>Therefore, QR<sup>2 <\/sup>= 169\u2212144<\/p>\n<p>QR<sup>2 <\/sup>= 25<\/p>\n<p>QR = \u221a25 = 5<\/p>\n<p>Therefore, the side QR = 5 cm<\/p>\n<p>To find tan P \u2013 cot R:<\/p>\n<p>According to the trigonometric ratio, the tangent function is equal to the ratio of the length of the opposite side to the adjacent sides, the value of tan (P) becomes<\/p>\n<p>tan (P) <strong>= <\/strong>Opposite side \/Adjacent side = QR\/PQ = 5\/12<\/p>\n<p>Since cot function is the reciprocal of the tan function, the ratio of cot function becomes,<\/p>\n<p>Cot (R) = Adjacent side\/Opposite side = QR\/PQ = 5\/12<\/p>\n<p>Therefore,<\/p>\n<p>tan (P) \u2013 cot (R) = 5\/12 \u2013 5\/12 = 0<\/p>\n<p>Therefore, tan(P) \u2013 cot(R) = 0<\/p>\n<p style=\"text-align: center;\"><strong>Also Check: <a href=\"https:\/\/infinitylearn.com\/surge\/study-material\/surface-areas-and-volumes\/class-10\/extra-questions\/maths\/chapter-13\/\">Surface Areas and Volumes Class 10 Extra Questions Maths Chapter 13<\/a><\/strong><\/p>\n<p><strong>3. If sin A = 3\/4, calculate cos A and tan A.<\/strong><\/p>\n<p>Solution:<\/p>\n<p>Let us assume a right angled triangle ABC, right angled at B<\/p>\n<p>Given: Sin A = 3\/4<\/p>\n<p>We know that, Sin function is the equal to the ratio of length of the opposite side to the hypotenuse side.<\/p>\n<p>Therefore, Sin A = Opposite side \/Hypotenuse= 3\/4<\/p>\n<p>Let BC be 3k and AC will be 4k<\/p>\n<p>where k is a positive real number.<\/p>\n<p>According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,<\/p>\n<p>AC<sup>2<\/sup>=AB<sup>2 <\/sup>+ BC<sup>2<\/sup><\/p>\n<p>Substitute the value of AC and BC<\/p>\n<p>(4k)<sup>2<\/sup>=AB<sup>2<\/sup> + (3k)<sup>2<\/sup><\/p>\n<p>16k<sup>2<\/sup>\u22129k<sup>2 <\/sup>=AB<sup>2<\/sup><\/p>\n<p>AB<sup>2<\/sup>=7k<sup>2<\/sup><\/p>\n<p>Therefore, AB = \u221a7k<\/p>\n<p>Now, we have to find the value of cos A and tan A<\/p>\n<p>We know that,<\/p>\n<p>Cos (A) = Adjacent side\/Hypotenuse<\/p>\n<p>Substitute the value of AB and AC and cancel the constant k in both numerator and denominator, we get<\/p>\n<p>AB\/AC = \u221a7k\/4k = \u221a7\/4<\/p>\n<p>Therefore, cos (A) = \u221a7\/4<\/p>\n<p>tan(A) = Opposite side\/Adjacent side<\/p>\n<p>Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get,<\/p>\n<p>BC\/AB = 3k\/\u221a7k = 3\/\u221a7<\/p>\n<p>Therefore, tan A = 3\/\u221a7<\/p>\n<p style=\"text-align: center;\"><strong>Also Check: <a href=\"https:\/\/infinitylearn.com\/surge\/study-material\/cbse-notes\/class-10\/maths\/polynomials-extra-questions-maths-chapter-2\/\" target=\"_blank\" rel=\"noopener\">Polynomials Class 10 Extra Questions Maths Chapter 2<\/a><\/strong><\/p>\n<p><strong>4. Given 15 cot A = 8, find sin A and sec A.<\/strong><\/p>\n<p>Solution:<\/p>\n<p>Let us assume a right angled triangle ABC, right angled at B<\/p>\n<p>Given: 15 cot A = 8<\/p>\n<p>So, Cot A = 8\/15<\/p>\n<p>We know that, cot function is the equal to the ratio of length of the adjacent side to the opposite side.<\/p>\n<p>Therefore, cot A = Adjacent side\/Opposite side = AB\/BC = 8\/15<\/p>\n<p>Let AB be 8k and BC will be 15k<\/p>\n<p>Where, k is a positive real number.<\/p>\n<p>According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,<\/p>\n<p>AC<sup>2<\/sup>=AB<sup>2 <\/sup>+ BC<sup>2<\/sup><\/p>\n<p>Substitute the value of AB and BC<\/p>\n<p>AC<sup>2<\/sup>= (8k)<sup>2<\/sup> + (15k)<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>= 64k<sup>2<\/sup> + 225k<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>= 289k<sup>2<\/sup><\/p>\n<p>Therefore, AC = 17k<\/p>\n<p>Now, we have to find the value of sin A and sec A<\/p>\n<p>We know that,<\/p>\n<p>Sin (A) = Opposite side \/Hypotenuse<\/p>\n<p>Substitute the value of BC and AC and cancel the constant k in both numerator and denominator, we get<\/p>\n<p>Sin A = BC\/AC = 15k\/17k = 15\/17<\/p>\n<p>Therefore, sin A = 15\/17<\/p>\n<p>Since secant or sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side.<\/p>\n<p>Sec (A) = Hypotenuse\/Adjacent side<\/p>\n<p>Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get,<\/p>\n<p>AC\/AB = 17k\/8k = 17\/8<\/p>\n<p>Therefore sec (A) = 17\/8<\/p>\n<p style=\"text-align: center;\"><strong>Also Check: <a href=\"https:\/\/infinitylearn.com\/surge\/study-material\/cbse-notes\/class-10\/maths\/some-applications-of-trigonometry-class-10-extra-questions-maths-chapter-9\/\" target=\"_blank\" rel=\"noopener\">Some Applications of Trigonometry Class 10 Extra Questions Maths Chapter 9<\/a><\/strong><\/p>\n<p><strong>5. Given sec \u03b8 = 13\/12 Calculate all other trigonometric ratios<\/strong><\/p>\n<p>Solution:<\/p>\n<p>We know that sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side<\/p>\n<p>Let us assume a right angled triangle ABC, right angled at B<\/p>\n<p>sec \u03b8 =13\/12 = Hypotenuse\/Adjacent side = AC\/AB<\/p>\n<p>Let AC be 13k and AB will be 12k<\/p>\n<p>Where, k is a positive real number.<\/p>\n<p>According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get,<\/p>\n<p>AC<sup>2<\/sup>=AB<sup>2 <\/sup>+ BC<sup>2<\/sup><\/p>\n<p>Substitute the value of AB and AC<\/p>\n<p>(13k)<sup>2<\/sup>= (12k)<sup>2<\/sup> + BC<sup>2<\/sup><\/p>\n<p>169k<sup>2<\/sup>= 144k<sup>2<\/sup> + BC<sup>2<\/sup><\/p>\n<p>169k<sup>2<\/sup>= 144k<sup>2<\/sup> + BC<sup>2<\/sup><\/p>\n<p>BC<sup>2 = <\/sup>169k<sup>2<\/sup> \u2013 144k<sup>2<\/sup><\/p>\n<p>BC<sup>2<\/sup>= 25k<sup>2<\/sup><\/p>\n<p>Therefore, BC = 5k<\/p>\n<p>Now, substitute the corresponding values in all other trigonometric ratios<\/p>\n<p>So,<\/p>\n<p>Sin \u03b8 = Opposite Side\/Hypotenuse = BC\/AC = 5\/13<\/p>\n<p>Cos \u03b8 = Adjacent Side\/Hypotenuse = AB\/AC = 12\/13<\/p>\n<p>tan \u03b8 = Opposite Side\/Adjacent Side = BC\/AB = 5\/12<\/p>\n<p>Cosec \u03b8 = Hypotenuse\/Opposite Side = AC\/BC = 13\/5<\/p>\n<p>cot \u03b8 = Adjacent Side\/Opposite Side = AB\/BC = 12\/5<\/p>\n<p><strong>6. If \u2220A and \u2220B are acute angles such that cos A = cos B, then show that \u2220 A = \u2220 B.<\/strong><\/p>\n<p>Solution:<\/p>\n<p>Let us assume the triangle ABC in which CD\u22a5AB<\/p>\n<p>Give that the angles A and B are acute angles, such that<\/p>\n<p>Cos (A) = cos (B)<\/p>\n<p>As per the angles taken, the cos ratio is written as<\/p>\n<p>AD\/AC = BD\/BC<\/p>\n<p>Now, interchange the terms, we get<\/p>\n<p>AD\/BD = AC\/BC<\/p>\n<p>Let take a constant value<\/p>\n<p>AD\/BD = AC\/BC = k<\/p>\n<p>Now consider the equation as<\/p>\n<p>AD = k BD \u2026(1)<\/p>\n<p>AC = k BC \u2026(2)<\/p>\n<p>By applying Pythagoras theorem in \u25b3CAD and \u25b3CBD we get,<\/p>\n<p>CD<sup>2<\/sup> = BC<sup>2<\/sup> \u2013 BD<sup>2 <\/sup>\u2026 (3)<\/p>\n<p>CD<sup>2 <\/sup>=AC<sup>2 <\/sup>\u2212AD<sup>2<\/sup> \u2026.(4)<\/p>\n<p>From the equations (3) and (4) we get,<\/p>\n<p>AC<sup>2<\/sup>\u2212AD<sup>2 <\/sup>= BC<sup>2<\/sup>\u2212BD<sup>2<\/sup><\/p>\n<p>Now substitute the equations (1) and (2) in (3) and (4)<\/p>\n<p>K<sup>2<\/sup>(BC<sup>2<\/sup>\u2212BD<sup>2<\/sup>)=(BC<sup>2<\/sup>\u2212BD<sup>2<\/sup>) k<sup>2<\/sup>=1<\/p>\n<p>Putting this value in equation, we obtain<\/p>\n<p>AC = BC<\/p>\n<p>\u2220A=\u2220B (Angles opposite to equal side are equal-isosceles triangle)<\/p>\n<p><strong>7. If cot \u03b8 = 7\/8, evaluate :<\/strong><\/p>\n<p><strong>(i) (1 + sin \u03b8)(1 \u2013 sin \u03b8)\/(1+cos \u03b8)(1-cos \u03b8)<\/strong><\/p>\n<p><strong>(ii) cot<sup>2<\/sup> \u03b8<\/strong><\/p>\n<p>Solution:<\/p>\n<p>Let us assume a \u25b3ABC in which \u2220B = 90\u00b0 and \u2220C = \u03b8<\/p>\n<p>Given:<\/p>\n<p>cot \u03b8 = BC\/AB = 7\/8<\/p>\n<p>Let BC = 7k and AB = 8k, where k is a positive real number<\/p>\n<p>According to Pythagoras theorem in \u25b3ABC we get.<\/p>\n<p>AC<sup>2 <\/sup>= AB<sup>2<\/sup>+BC<sup>2<\/sup><\/p>\n<p>AC<sup>2 <\/sup>= (8k)<sup>2<\/sup>+(7k)<sup>2<\/sup><\/p>\n<p>AC<sup>2 <\/sup>= 64k<sup>2<\/sup>+49k<sup>2<\/sup><\/p>\n<p>AC<sup>2 <\/sup>= 113k<sup>2<\/sup><\/p>\n<p>AC = \u221a113 k<\/p>\n<p>According to the sine and cos function ratios, it is written as<\/p>\n<p>sin \u03b8 = AB\/AC = Opposite Side\/Hypotenuse = 8k\/\u221a113 k = 8\/\u221a113 and<\/p>\n<p>cos \u03b8 = Adjacent Side\/Hypotenuse = BC\/AC = 7k\/\u221a113 k = 7\/\u221a113<\/p>\n<p>Now apply the values of sin function and cos function:<\/p>\n<p><img loading=\"lazy\" class=\"alignnone size-full wp-image-711438\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2022\/01\/Screenshot-2024-03-07-124939.png\" alt=\"\" width=\"283\" height=\"205\" \/><\/p>\n<p><strong>8. If 3 cot A = 4, check whether (1-tan<sup>2 <\/sup>A)\/(1+tan<sup>2<\/sup> A) = cos<sup>2<\/sup> A \u2013 sin <sup>2 <\/sup>A or not.<\/strong><\/p>\n<p>Solution:<\/p>\n<p>Let \u25b3ABC in which \u2220B=90\u00b0<\/p>\n<p>We know that, cot function is the reciprocal of tan function and it is written as<\/p>\n<p>cot(A) = AB\/BC = 4\/3<\/p>\n<p>Let AB = 4k an BC =3k, where k is a positive real number.<\/p>\n<p>According to the Pythagorean theorem,<\/p>\n<p>AC<sup>2<\/sup>=AB<sup>2<\/sup>+BC<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>=(4k)<sup>2<\/sup>+(3k)<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>=16k<sup>2<\/sup>+9k<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>=25k<sup>2<\/sup><\/p>\n<p>AC=5k<\/p>\n<p>Now, apply the values corresponding to the ratios<\/p>\n<p>tan(A) = BC\/AB = 3\/4<\/p>\n<p>sin (A) = BC\/AC = 3\/5<\/p>\n<p>cos (A) = AB\/AC = 4\/5<\/p>\n<p>Now compare the left hand side(LHS) with right hand side(RHS)<\/p>\n<p><img loading=\"lazy\" class=\"alignnone size-medium wp-image-711440\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2022\/01\/Screenshot-2024-03-07-125030-300x106.png\" alt=\"\" width=\"300\" height=\"106\" srcset=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2022\/01\/Screenshot-2024-03-07-125030-300x106.png 300w, https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2022\/01\/Screenshot-2024-03-07-125030.png 499w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>Since, both the LHS and RHS = 7\/25<\/p>\n<p>R.H.S. =L.H.S.<\/p>\n<p>Hence, <strong>(1-tan<sup>2 <\/sup>A)\/(1+tan<sup>2<\/sup> A) = cos<sup>2<\/sup> A \u2013 sin <sup>2 <\/sup>A<\/strong>  is proved<\/p>\n<p><strong>9. In triangle ABC, right-angled at B, if tan A = 1\/\u221a3 find the value of:<\/strong><\/p>\n<p><strong>(i) sin A cos C + cos A sin C<\/strong><\/p>\n<p><strong>(ii) cos A cos C \u2013 sin A sin C<\/strong><\/p>\n<p>Solution:<\/p>\n<p>Let \u0394ABC in which \u2220B=90\u00b0<\/p>\n<p>tan A = BC\/AB = 1\/\u221a3<\/p>\n<p>Let BC = 1k and AB = \u221a3 k,<\/p>\n<p>Where k is the positive real number of the problem<\/p>\n<p>By Pythagoras theorem in \u0394ABC we get:<\/p>\n<p>AC<sup>2<\/sup>=AB<sup>2<\/sup>+BC<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>=(\u221a3 k)<sup>2<\/sup>+(k)<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>=3k<sup>2<\/sup>+k<sup>2<\/sup><\/p>\n<p>AC<sup>2<\/sup>=4k<sup>2<\/sup><\/p>\n<p>AC = 2k<\/p>\n<p>Now find the values of cos A, Sin A<\/p>\n<p>Sin A = BC\/AC = 1\/2<\/p>\n<p>Cos A = AB\/AC = \u221a3\/2<\/p>\n<p>Then find the values of cos C and sin C<\/p>\n<p>Sin C = AB\/AC = <strong>\u221a<\/strong>3\/2<\/p>\n<p>Cos C = BC\/AC = 1\/2<\/p>\n<p>Now, substitute the values in the given problem<\/p>\n<p>(i) sin A cos C + cos A sin C = (1\/2) \u00d7(1\/2 )+ \u221a3\/2 \u00d7\u221a3\/2 = 1\/4 + 3\/4 = 1<\/p>\n<p>(ii) cos A cos C \u2013 sin A sin C = (<strong>\u221a<\/strong>3\/2 )(1\/2) \u2013 (1\/2) (<strong>\u221a<\/strong>3\/2 ) = 0<\/p>\n<p><strong>10. In \u2206 PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P<\/strong><\/p>\n<p>Solution:<\/p>\n<p>In a given triangle PQR, right angled at Q, the following measures are<\/p>\n<p>PQ = 5 cm<\/p>\n<p>PR + QR = 25 cm<\/p>\n<p>Now let us assume, QR = x<\/p>\n<p>PR = 25-QR<\/p>\n<p>PR = 25- x<\/p>\n<p>According to the Pythagorean Theorem,<\/p>\n<p>PR<sup>2<\/sup> = PQ<sup>2<\/sup> + QR<sup>2<\/sup><\/p>\n<p>Substitute the value of PR as x<\/p>\n<p>(25- x)<sup> 2 <\/sup>= 5<sup>2 <\/sup>+ x<sup>2<\/sup><\/p>\n<p>25<sup>2<\/sup> + x<sup>2<\/sup> \u2013 50x = 25 + x<sup>2<\/sup><\/p>\n<p>625 + x<sup>2<\/sup>-50x -25 \u2013 x<sup>2 <\/sup>= 0<\/p>\n<p>-50x = -600<\/p>\n<p>x= -600\/-50<\/p>\n<p>x = 12 = QR<\/p>\n<p>Now, find the value of PR<\/p>\n<p>PR = 25- QR<\/p>\n<p>Substitute the value of QR<\/p>\n<p>PR = 25-12<\/p>\n<p>PR = 13<\/p>\n<p>Now, substitute the value to the given problem<\/p>\n<p>(1) sin p = Opposite Side\/Hypotenuse = QR\/PR = 12\/13<\/p>\n<p>(2) Cos p = Adjacent Side\/Hypotenuse = PQ\/PR = 5\/13<\/p>\n<p>(3) tan p =Opposite Side\/Adjacent side = QR\/PQ = 12\/5<\/p>\n<p><strong>11. State whether the following are true or false. Justify your answer.<\/strong><\/p>\n<p><strong>(i) The value of tan A is always less than 1.<\/strong><\/p>\n<p><strong>(ii) sec A = 12\/5 for some value of angle A.<\/strong><\/p>\n<p><strong>(iii)cos A is the abbreviation used for the cosecant of angle A.<\/strong><\/p>\n<p><strong>(iv) cot A is the product of cot and A.<\/strong><\/p>\n<p><strong>(v) sin \u03b8 = 4\/3 for some angle \u03b8.<\/strong><\/p>\n<p>Solution:<\/p>\n<p><strong>(i) <\/strong>The value of tan A is always less than 1.<\/p>\n<p>Answer: <strong>False<\/strong><\/p>\n<p>Proof: In \u0394MNC in which \u2220N = 90\u2218,<\/p>\n<p>MN = 3, NC = 4 and MC = 5<\/p>\n<p>Value of tan M = 4\/3 which is greater than 1.<\/p>\n<p>The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as it will follow the Pythagoras theorem.<\/p>\n<p>MC<sup>2<\/sup>=MN<sup>2<\/sup>+NC<sup>2<\/sup><\/p>\n<p>5<sup>2<\/sup>=3<sup>2<\/sup>+4<sup>2<\/sup><\/p>\n<p>25=9+16<\/p>\n<p>25<sup> <\/sup>=<sup> <\/sup>25<\/p>\n<p><strong>(ii)<\/strong> sec A = 12\/5 for some value of angle A<\/p>\n<p>Answer: <strong>True<\/strong><\/p>\n<p>Justification: Let a \u0394MNC in which \u2220N = 90\u00ba,<\/p>\n<p>MC=12k and MB=5k, where k is a positive real number.<\/p>\n<p>By Pythagoras theorem we get,<\/p>\n<p>MC<sup>2<\/sup>=MN<sup>2<\/sup>+NC<sup>2<\/sup><\/p>\n<p>(12k)<sup>2<\/sup>=(5k)<sup>2<\/sup>+NC<sup>2<\/sup><\/p>\n<p>NC<sup>2<\/sup>+25k<sup>2<\/sup>=144k<sup>2<\/sup><\/p>\n<p>NC<sup>2<\/sup>=119k<sup>2<\/sup><\/p>\n<p>Such a triangle is possible as it will follow the Pythagoras theorem.<\/p>\n<p><strong>(iii)<\/strong> cos A is the abbreviation used for the cosecant of angle A.<\/p>\n<p>Answer: <strong>False<\/strong><\/p>\n<p>Justification: Abbreviation used for cosecant of angle M is cosec M. cos M is the abbreviation used for cosine of angle M.<\/p>\n<p><strong>(iv)<\/strong> cot A is the product of cot and A.<\/p>\n<p>Answer:<strong> False<\/strong><\/p>\n<p>Justification: cot M is not the product of cot and M. It is the cotangent of \u2220M.<\/p>\n<p><strong>(v)<\/strong> sin \u03b8 = 4\/3 for some angle \u03b8.<\/p>\n<p>Answer<strong>: False<\/strong><\/p>\n<p>Justification: sin \u03b8 = Opposite\/Hypotenuse<\/p>\n<p>We know that in a right angled triangle, Hypotenuse is the longest side.<\/p>\n<p>\u2234 sin \u03b8 will always less than 1 and it can never be 4\/3 for any value of \u03b8.<\/p>\n<hr \/>\n<p><strong>Exercise 8.2 Page: 187<\/strong><\/p>\n<p><strong>1. Evaluate the following:<\/strong><\/p>\n<p><strong>(i) sin 60\u00b0 cos 30\u00b0 + sin 30\u00b0 cos 60\u00b0<\/strong><\/p>\n<p><strong>(ii) 2 tan<sup>2<\/sup> 45\u00b0 + cos<sup>2<\/sup> 30\u00b0 \u2013 sin<sup>2<\/sup> 60<\/strong><\/p>\n<p><img loading=\"lazy\" class=\"alignnone size-medium wp-image-711456\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2022\/01\/Screenshot-2024-03-07-130215-300x163.png\" alt=\"\" width=\"300\" height=\"163\" srcset=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2022\/01\/Screenshot-2024-03-07-130215-300x163.png 300w, https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2022\/01\/Screenshot-2024-03-07-130215.png 409w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>Solution:<\/p>\n<p>(i) sin 60\u00b0 cos 30\u00b0 + sin 30\u00b0 cos 60\u00b0<\/p>\n<p>First, find the values of the given trigonometric ratios<\/p>\n<p>sin 30\u00b0 = 1\/2<\/p>\n<p>cos 30\u00b0 = \u221a3\/2<\/p>\n<p>sin 60\u00b0 = 3\/2<\/p>\n<p>cos 60\u00b0= 1\/2<\/p>\n<p>Now, substitute the values in the given problem<\/p>\n<p>sin 60\u00b0 cos 30\u00b0 + sin 30\u00b0 cos 60\u00b0 = \u221a3\/2 \u00d7\u221a3\/2 + (1\/2) \u00d7(1\/2 ) = 3\/4+1\/4 = 4\/4 =1<\/p>\n<p>(ii) 2 tan<sup>2<\/sup> 45\u00b0 + cos<sup>2<\/sup> 30\u00b0 \u2013 sin<sup>2<\/sup> 60<\/p>\n<p>We know that, the values of the trigonometric ratios are:<\/p>\n<p>sin 60\u00b0 = \u221a3\/2<\/p>\n<p>cos 30\u00b0 = \u221a3\/2<\/p>\n<p>tan 45\u00b0 = 1<\/p>\n<p>Substitute the values in the given problem<\/p>\n<p>2 tan<sup>2<\/sup> 45\u00b0 + cos<sup>2<\/sup> 30\u00b0 \u2013 sin<sup>2<\/sup> 60 = 2(1)<sup>2 <\/sup>+ (\u221a3\/2)<sup>2<\/sup>-(\u221a3\/2)<sup>2<\/sup><\/p>\n<p>2 tan<sup>2<\/sup> 45\u00b0 + cos<sup>2<\/sup> 30\u00b0 \u2013 sin<sup>2<\/sup> 60 = 2 + 0<\/p>\n<p>2 tan<sup>2<\/sup> 45\u00b0 + cos<sup>2<\/sup> 30\u00b0 \u2013 sin<sup>2<\/sup> 60 = 2<\/p>\n<p>(iii) cos 45\u00b0\/(sec 30\u00b0+cosec 30\u00b0)<\/p>\n<p>We know that,<\/p>\n<p>cos 45\u00b0 = 1\/\u221a2<\/p>\n<p>sec 30\u00b0 = 2\/\u221a3<\/p>\n<p>cosec 30\u00b0 = 2<\/p>\n<p>Substitute the values, we get<\/p>\n<p><img loading=\"lazy\" class=\"alignnone size-medium wp-image-711436\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2022\/01\/Screenshot-2024-03-07-124119-300x236.png\" alt=\"\" width=\"300\" height=\"236\" srcset=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2022\/01\/Screenshot-2024-03-07-124119-300x236.png 300w, https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2022\/01\/Screenshot-2024-03-07-124119.png 673w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>We know that,<\/p>\n<p>sin 30\u00b0 = 1\/2<\/p>\n<p>tan 45\u00b0 = 1<\/p>\n<p>cosec 60\u00b0 = 2\/\u221a3<\/p>\n<p>sec 30\u00b0 = 2\/\u221a3<\/p>\n<p>cos 60\u00b0 = 1\/2<\/p>\n<p>cot 45\u00b0 = 1<\/p>\n<p>Substitute the values in the given problem, we get<\/p>\n<p><img loading=\"lazy\" class=\"alignnone size-medium wp-image-711457\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2022\/01\/Screenshot-2024-03-07-130325-300x226.png\" alt=\"\" width=\"300\" height=\"226\" srcset=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2022\/01\/Screenshot-2024-03-07-130325-300x226.png 300w, https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2022\/01\/Screenshot-2024-03-07-130325.png 601w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>We know that,<\/p>\n<p>cos 60\u00b0 = 1\/2<\/p>\n<p>sec 30\u00b0 = 2\/\u221a3<\/p>\n<p>tan 45\u00b0 = 1<\/p>\n<p>sin 30\u00b0 = 1\/2<\/p>\n<p>cos 30\u00b0 = \u221a3\/2<\/p>\n<p>Now, substitute the values in the given problem, we get<\/p>\n<p>(5cos<sup>2<\/sup>60\u00b0 + 4sec<sup>2<\/sup>30\u00b0 \u2013 tan<sup>2<\/sup>45\u00b0)\/(sin<sup>2 <\/sup>30\u00b0 + cos<sup>2 <\/sup>30\u00b0)<\/p>\n<p>= 5(1\/2)<sup>2<\/sup>+4(2\/\u221a3)<sup>2<\/sup>-1<sup>2<\/sup>\/(1\/2)<sup>2<\/sup>+(\u221a3\/2)<sup>2<\/sup><\/p>\n<p><sup> <\/sup>= (5\/4+16\/3-1)\/(1\/4+3\/4)<\/p>\n<p>= (15+64-12)\/12\/(4\/4)<\/p>\n<p>= 67\/12<\/p>\n<p><strong>2. Choose the correct option and justify your choice :<br \/>\n(i) 2tan 30\u00b0\/1+tan<sup>2<\/sup>30\u00b0 =<br \/>\n(A) sin 60\u00b0            (B) cos 60\u00b0          (C) tan 60\u00b0            (D) sin 30\u00b0<br \/>\n(ii) 1-tan<sup>2<\/sup>45\u00b0\/1+tan<sup>2<\/sup>45\u00b0 =<br \/>\n(A) tan 90\u00b0            (B) 1                    (C) sin 45\u00b0            (D) 0<br \/>\n(iii)  sin 2A = 2 sin A is true when A =<br \/>\n(A) 0\u00b0                   (B) 30\u00b0                  (C) 45\u00b0                 (D) 60\u00b0<\/strong><\/p>\n<p><strong>(iv) 2tan30\u00b0\/1-tan<sup>2<\/sup>30\u00b0 =<br \/>\n(A) cos 60\u00b0          (B) sin 60\u00b0             (C) tan 60\u00b0           (D) sin 30\u00b0<\/strong><\/p>\n<p>Solution:<\/p>\n<p>(i) (A) is correct.<\/p>\n<p>Substitute the of tan 30\u00b0 in the given equation<\/p>\n<p>tan 30\u00b0 = 1\/\u221a3<\/p>\n<p>2tan 30\u00b0\/1+tan<sup>2<\/sup>30\u00b0 = 2(1\/\u221a3)\/1+(1\/\u221a3)<sup>2<\/sup><\/p>\n<p>= (2\/\u221a3)\/(1+1\/3) = (2\/\u221a3)\/(4\/3)<\/p>\n<p>= 6\/4\u221a3 = \u221a3\/2 = sin 60\u00b0<\/p>\n<p>The obtained solution is equivalent to the trigonometric ratio sin 60\u00b0<\/p>\n<p>(ii) (D) is correct.<\/p>\n<p>Substitute the of tan 45\u00b0 in the given equation<\/p>\n<p>tan 45\u00b0 = 1<\/p>\n<p>1-tan<sup>2<\/sup>45\u00b0\/1+tan<sup>2<\/sup>45\u00b0 = (1-1<sup>2<\/sup>)\/(1+1<sup>2<\/sup>)<\/p>\n<p>= 0\/2 = 0<\/p>\n<p>The solution of the above equation is 0.<\/p>\n<p>(iii) (A) is correct.<\/p>\n<p>To find the value of A, substitute the degree given in the options one by one<\/p>\n<p>sin 2A = 2 sin A is true when A = 0\u00b0<\/p>\n<p>As sin 2A = sin 0\u00b0 = 0<\/p>\n<p>2 sin A = 2 sin 0\u00b0 = 2 \u00d7 0 = 0<\/p>\n<p>or,<\/p>\n<p>Apply the sin 2A formula, to find the degree value<\/p>\n<p>sin 2A = 2sin A cos A<\/p>\n<p>\u21d22sin A cos A = 2 sin A<\/p>\n<p>\u21d2 2cos A = 2 \u21d2 cos A = 1<\/p>\n<p>Now, we have to check, to get the solution as 1, which degree value has to be applied.<\/p>\n<p>When 0 degree is applied to cos value, i.e., cos 0 =1<\/p>\n<p>Therefore, \u21d2 A = 0\u00b0<\/p>\n<p>(iv) (C) is correct.<\/p>\n<p>Substitute the of tan 30\u00b0 in the given equation<\/p>\n<p>tan 30\u00b0 = 1\/\u221a3<\/p>\n<p>2tan30\u00b0\/1-tan<sup>2<\/sup>30\u00b0 =  2(1\/\u221a3)\/1-(1\/\u221a3)<sup>2<\/sup><\/p>\n<p>= (2\/\u221a3)\/(1-1\/3) = (2\/\u221a3)\/(2\/3) = \u221a3 = tan 60\u00b0<\/p>\n<p>The value of the given equation is equivalent to tan 60\u00b0.<\/p>\n<p><strong>3. If tan (A + B) = \u221a3 and tan (A \u2013 B) = 1\/\u221a3 ,0\u00b0 &lt; A + B \u2264 90\u00b0; A &gt; B, find A and B.<\/strong><\/p>\n<p>Solution:<\/p>\n<p>tan (A + B) = \u221a3<\/p>\n<p>Since \u221a3 = tan 60\u00b0<\/p>\n<p>Now substitute the degree value<\/p>\n<p>\u21d2 tan (A + B) = tan 60\u00b0<\/p>\n<p>(A + B) = 60\u00b0 \u2026 (i)<\/p>\n<p>The above equation is assumed as equation (i)<\/p>\n<p>tan (A \u2013 B) = 1\/\u221a3<\/p>\n<p>Since 1\/\u221a3 = tan 30\u00b0<\/p>\n<p>Now substitute the degree value<\/p>\n<p>\u21d2 tan (A \u2013 B) = tan 30\u00b0<\/p>\n<p>(A \u2013 B) = 30\u00b0 \u2026 equation (ii)<\/p>\n<p>Now add the equation (i) and (ii), we get<\/p>\n<p>A + B + A \u2013 B = 60\u00b0 + 30\u00b0<\/p>\n<p>Cancel the terms B<\/p>\n<p>2A = 90\u00b0<\/p>\n<p>A= 45\u00b0<\/p>\n<p>Now, substitute the value of A in equation (i) to find the value of B<\/p>\n<p>45\u00b0 + B = 60\u00b0<\/p>\n<p>B = 60\u00b0 \u2013 45\u00b0<\/p>\n<p>B = 15\u00b0<\/p>\n<p>Therefore A = 45\u00b0 and B = 15\u00b0<\/p>\n<p><strong>4. State whether the following are true or false. Justify your answer.<\/strong><\/p>\n<p><strong>(i) sin (A + B) = sin A + sin B.<\/strong><\/p>\n<p><strong>(ii) The value of sin \u03b8 increases as \u03b8 increases.<\/strong><\/p>\n<p><strong>(iii) The value of cos \u03b8 increases as \u03b8 increases.<\/strong><\/p>\n<p><strong>(iv) sin \u03b8 = cos \u03b8 for all values of \u03b8.<\/strong><\/p>\n<p><strong>(v) cot A is not defined for A = 0\u00b0.<\/strong><\/p>\n<p>Solution:<\/p>\n<p>(i) False.<\/p>\n<p>Justification:<\/p>\n<p>Let us take A = 30\u00b0 and B = 60\u00b0, then<\/p>\n<p>Substitute the values in the sin (A + B) formula, we get<\/p>\n<p>sin (A + B) = sin (30\u00b0 + 60\u00b0) = sin 90\u00b0 = 1 and,<\/p>\n<p>sin A + sin B = sin 30\u00b0 + sin 60\u00b0<\/p>\n<p>= 1\/2 + \u221a3\/2 = 1+\u221a3\/2<\/p>\n<p>Since the values obtained are not equal, the solution is false.<\/p>\n<p>(ii) True.<\/p>\n<p>Justification:<\/p>\n<p>According to the values obtained as per the unit circle, the values of sin are:<\/p>\n<p>sin 0\u00b0 = 0<\/p>\n<p>sin 30\u00b0 = 1\/2<\/p>\n<p>sin 45\u00b0 = 1\/\u221a2<\/p>\n<p>sin 60\u00b0 = \u221a3\/2<\/p>\n<p>sin 90\u00b0 = 1<\/p>\n<p>Thus the value of sin \u03b8 increases as \u03b8 increases. Hence, the statement is true<\/p>\n<p>(iii) False.<\/p>\n<p>According to the values obtained as per the unit circle, the values of cos are:<\/p>\n<p>cos 0\u00b0 = 1<\/p>\n<p>cos 30\u00b0 = \u221a3\/2<\/p>\n<p>cos 45\u00b0 = 1\/\u221a2<\/p>\n<p>cos 60\u00b0 = 1\/2<\/p>\n<p>cos 90\u00b0 = 0<\/p>\n<p>Thus, the value of cos \u03b8 decreases as \u03b8 increases. So, the statement given above is false.<\/p>\n<p>(iv) False<\/p>\n<p>sin \u03b8 = cos \u03b8, when a right triangle has 2 angles of (\u03c0\/4). Therefore, the above statement is false.<\/p>\n<p>(v) True.<\/p>\n<p>Since cot function is the reciprocal of the tan function, it is also written as:<\/p>\n<p>cot A = cos A\/sin A<\/p>\n<p>Now substitute A = 0\u00b0<\/p>\n<p>cot 0\u00b0 = cos 0\u00b0\/sin 0\u00b0 = 1\/0 = undefined.<\/p>\n<p>Hence, it is true<\/p>\n<p><strong>Exercise 8.3 Page: 189<\/strong><\/p>\n<p><strong>1. Evaluate :<\/strong><\/p>\n<p><strong>(i) sin 18\u00b0\/cos 72\u00b0        <\/strong><\/p>\n<p><strong>(ii) tan 26\u00b0\/cot 64\u00b0      <\/strong><\/p>\n<p><strong>(iii)  cos 48\u00b0 \u2013 sin 42\u00b0      <\/strong><\/p>\n<p><strong>(iv)  cosec 31\u00b0 \u2013 sec 59\u00b0<\/strong><\/p>\n<p>Solution:<\/p>\n<p>(i) sin 18\u00b0\/cos 72\u00b0<\/p>\n<p>To simplify this, convert the sin function into cos function<\/p>\n<p>We know that, 18\u00b0 is written as 90\u00b0 \u2013 18\u00b0, which is equal to the cos 72\u00b0.<\/p>\n<p>= sin (90\u00b0 \u2013 18\u00b0) \/cos 72\u00b0<\/p>\n<p>Substitute the value, to simplify this equation<\/p>\n<p>= cos 72\u00b0 \/cos 72\u00b0 = 1<\/p>\n<p>(ii) tan 26\u00b0\/cot 64\u00b0<\/p>\n<p>To simplify this, convert the tan function into cot function<\/p>\n<p>We know that, 26\u00b0 is written as 90\u00b0 \u2013 26\u00b0, which is equal to the cot 64\u00b0.<\/p>\n<p>= tan (90\u00b0 \u2013 26\u00b0)\/cot 64\u00b0<\/p>\n<p>Substitute the value, to simplify this equation<\/p>\n<p>= cot 64\u00b0\/cot 64\u00b0 = 1<\/p>\n<p>(iii) cos 48\u00b0 \u2013 sin 42\u00b0<\/p>\n<p>To simplify this, convert the cos function into sin function<\/p>\n<p>We know that, 48\u00b0 is written as 90\u00b0 \u2013 42\u00b0, which is equal to the sin 42\u00b0.<\/p>\n<p>= cos (90\u00b0 \u2013 42\u00b0) \u2013 sin 42\u00b0<\/p>\n<p>Substitute the value, to simplify this equation<\/p>\n<p>= sin 42\u00b0 \u2013 sin 42\u00b0 = 0<\/p>\n<p>(iv) cosec 31\u00b0 \u2013 sec 59\u00b0<\/p>\n<p>To simplify this, convert the cosec function into sec function<\/p>\n<p>We know that, 31\u00b0 is written as 90\u00b0 \u2013 59\u00b0, which is equal to the sec 59\u00b0<\/p>\n<p>= cosec (90\u00b0 \u2013 59\u00b0) \u2013 sec 59\u00b0<\/p>\n<p>Substitute the value, to simplify this equation<\/p>\n<p>= sec 59\u00b0 \u2013 sec 59\u00b0 = 0<\/p>\n<p><strong>2.  Show that:<\/strong><\/p>\n<p><strong>(i) tan 48\u00b0 tan 23\u00b0 tan 42\u00b0 tan 67\u00b0 = 1<\/strong><\/p>\n<p><strong>(ii) cos 38\u00b0 cos 52\u00b0 \u2013 sin 38\u00b0 sin 52\u00b0 = 0<\/strong><\/p>\n<p>Solution:<\/p>\n<p>(i) tan 48\u00b0 tan 23\u00b0 tan 42\u00b0 tan 67\u00b0<\/p>\n<p>Simplify the given problem by converting some of the tan functions to the cot functions<\/p>\n<p>We know that, tan 48\u00b0 = tan (90\u00b0 \u2013 42\u00b0) = cot 42\u00b0<\/p>\n<p>tan 23\u00b0 = tan (90\u00b0 \u2013 67\u00b0) = cot 67\u00b0<\/p>\n<p>= tan (90\u00b0 \u2013 42\u00b0) tan (90\u00b0 \u2013 67\u00b0) tan 42\u00b0 tan 67\u00b0<\/p>\n<p>Substitute the values<\/p>\n<p>= cot 42\u00b0 cot 67\u00b0 tan 42\u00b0 tan 67\u00b0<\/p>\n<p>= (cot 42\u00b0 tan 42\u00b0) (cot 67\u00b0 tan 67\u00b0) = 1\u00d71 = 1<\/p>\n<p>(ii) cos 38\u00b0 cos 52\u00b0 \u2013 sin 38\u00b0 sin 52\u00b0<\/p>\n<p>Simplify the given problem by converting some of the cos functions to the sin functions<\/p>\n<p>We know that,<\/p>\n<p>cos 38\u00b0 = cos (90\u00b0 \u2013 52\u00b0) = sin 52\u00b0<\/p>\n<p>cos 52\u00b0= cos (90\u00b0-38\u00b0) = sin 38\u00b0<\/p>\n<p>= cos (90\u00b0 \u2013 52\u00b0) cos (90\u00b0-38\u00b0) \u2013 sin 38\u00b0 sin 52\u00b0<\/p>\n<p>Substitute the values<\/p>\n<p>= sin 52\u00b0 sin 38\u00b0 \u2013 sin 38\u00b0 sin 52\u00b0 = 0<\/p>\n<p><strong>3. If tan 2A = cot (A \u2013 18\u00b0), where 2A is an acute angle, find the value of A<\/strong>.<\/p>\n<p>Solution:<\/p>\n<p>tan 2A = cot (A- 18\u00b0)<\/p>\n<p>We know that tan 2A = cot (90\u00b0 \u2013 2A)<\/p>\n<p>Substitute the above equation in the given problem<\/p>\n<p>\u21d2 cot (90\u00b0 \u2013 2A) = cot (A -18\u00b0)<\/p>\n<p>Now, equate the angles,<\/p>\n<p>\u21d2 90\u00b0 \u2013 2A = A- 18\u00b0 \u21d2 108\u00b0 = 3A<\/p>\n<p>A = 108\u00b0 \/ 3<\/p>\n<p>Therefore, the value of A = 36\u00b0<\/p>\n<p><strong>4.  If tan A = cot B, prove that A + B = 90\u00b0.<\/strong><\/p>\n<p>Solution:<\/p>\n<p>tan A = cot B<\/p>\n<p>We know that cot B = tan (90\u00b0 \u2013 B)<\/p>\n<p>To prove A + B = 90\u00b0, substitute the above equation in the given problem<\/p>\n<p>tan A = tan (90\u00b0 \u2013 B)<\/p>\n<p>A = 90\u00b0 \u2013 B<\/p>\n<p>A + B = 90\u00b0<\/p>\n<p>Hence Proved.<\/p>\n<p><strong>5. If sec 4A = cosec (A \u2013 20\u00b0), where 4A is an acute angle, find the value of A.<\/strong><\/p>\n<p>Solution:<\/p>\n<p>sec 4A = cosec (A \u2013 20\u00b0)<\/p>\n<p>We know that sec 4A = cosec (90\u00b0 \u2013 4A)<\/p>\n<p>To find the value of A, substitute the above equation in the given problem<\/p>\n<p>cosec (90\u00b0 \u2013 4A) = cosec (A \u2013 20\u00b0)<\/p>\n<p>Now, equate the angles<\/p>\n<p>90\u00b0 \u2013 4A= A- 20\u00b0<\/p>\n<p>110\u00b0 = 5A<\/p>\n<p>A = 110\u00b0\/ 5 = 22\u00b0<\/p>\n<p>Therefore, the value of A = 22\u00b0<\/p>\n<p><strong>6. If A, B and C are interior angles of a triangle ABC, then show that<\/strong><\/p>\n<p><strong>    sin (B+C\/2) = cos A\/2<\/strong><\/p>\n<p>Solution:<\/p>\n<p>We know that, for a given triangle, sum of all the interior angles of a triangle is equal to 180\u00b0<\/p>\n<p>A + B + C = 180\u00b0 \u2026.(1)<\/p>\n<p>To find the value of (B+ C)\/2, simplify the equation (1)<\/p>\n<p>\u21d2 B + C = 180\u00b0 \u2013 A<\/p>\n<p>\u21d2 (B+C)\/2 = (180\u00b0-A)\/2<\/p>\n<p>\u21d2 (B+C)\/2 = (90\u00b0-A\/2)<\/p>\n<p>Now, multiply both sides by sin functions, we get<\/p>\n<p>\u21d2 sin (B+C)\/2 = sin (90\u00b0-A\/2)<\/p>\n<p>Since sin (90\u00b0-A\/2) = cos A\/2, the above equation is equal to<\/p>\n<p>sin (B+C)\/2 = cos A\/2<\/p>\n<p>Hence proved.<\/p>\n<p><strong>7. Express sin 67\u00b0 + cos 75\u00b0 in terms of trigonometric ratios of angles between 0\u00b0 and 45\u00b0.<\/strong><\/p>\n<p>Solution:<\/p>\n<p>Given:<\/p>\n<p>sin 67\u00b0 + cos 75\u00b0<\/p>\n<p>In term of sin as cos function and cos as sin function, it can be written as follows<\/p>\n<p>sin 67\u00b0 = sin (90\u00b0 \u2013 23\u00b0)<\/p>\n<p>cos 75\u00b0 = cos (90\u00b0 \u2013 15\u00b0)<\/p>\n<p>So, sin 67\u00b0 + cos 75\u00b0 = sin (90\u00b0 \u2013 23\u00b0) + cos (90\u00b0 \u2013 15\u00b0)<\/p>\n<p>Now, simplify the above equation<\/p>\n<p>= cos 23\u00b0 + sin 15\u00b0<\/p>\n<p>Therefore, sin 67\u00b0 + cos 75\u00b0 is also expressed as cos 23\u00b0 + sin 15\u00b0<\/p>\n<p><strong>Exercise 8.4 Page: 193<\/strong><\/p>\n<p><strong>1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.<\/strong><\/p>\n<p>Solution:<\/p>\n<p>To convert the given trigonometric ratios in terms of cot functions, use trigonometric formulas<\/p>\n<p>We know that,<\/p>\n<p>cosec<sup>2<\/sup>A<sup> <\/sup>\u2013 cot<sup>2<\/sup>A = 1<\/p>\n<p>cosec<sup>2<\/sup>A = 1 + cot<sup>2<\/sup>A<\/p>\n<p>Since cosec function is the inverse of sin function, it is written as<\/p>\n<p>1\/sin<sup>2<\/sup>A = 1 + cot<sup>2<\/sup>A<\/p>\n<p>Now, rearrange the terms, it becomes<\/p>\n<p>sin<sup>2<\/sup>A = 1\/(1+cot<sup>2<\/sup>A)<\/p>\n<p>Now, take square roots on both sides, we get<\/p>\n<p>sin A = \u00b11\/(\u221a(1+cot<sup>2<\/sup>A)<\/p>\n<p>The above equation defines the sin function in terms of cot function<\/p>\n<p>Now, to express sec function in terms of cot function, use this formula<\/p>\n<p>sin<sup>2<\/sup>A = 1\/ (1+cot<sup>2<\/sup>A)<\/p>\n<p>Now, represent the sin function as cos function<\/p>\n<p>1 \u2013 cos<sup>2<\/sup>A = 1\/ (1+cot<sup>2<\/sup>A)<\/p>\n<p>Rearrange the terms,<\/p>\n<p>cos<sup>2<\/sup>A = 1 \u2013 1\/(1+cot<sup>2<\/sup>A)<\/p>\n<p>\u21d2cos<sup>2<\/sup>A = (1-1+cot<sup>2<\/sup>A)\/(1+cot<sup>2<\/sup>A)<\/p>\n<p>Since sec function is the inverse of cos function,<\/p>\n<p>\u21d2 1\/sec<sup>2<\/sup>A = cot<sup>2<\/sup>A\/(1+cot<sup>2<\/sup>A)<\/p>\n<p>Take the reciprocal and square roots on both sides, we get<\/p>\n<p>\u21d2 sec A = \u00b1\u221a (1+cot<sup>2<\/sup>A)\/cotA<\/p>\n<p>Now, to express tan function in terms of cot function<\/p>\n<p>tan A = sin A\/cos A and cot A = cos A\/sin A<\/p>\n<p>Since cot function is the inverse of tan function, it is rewritten as<\/p>\n<p>tan A = 1\/cot A<\/p>\n<p><strong>2. Write all the other trigonometric ratios of \u2220A in terms of sec A.<br \/>\n<\/strong><br \/>\nSolution:<\/p>\n<p>Cos A function in terms of sec A:<\/p>\n<p>sec A = 1\/cos A<\/p>\n<p>\u21d2 cos A = 1\/sec A<\/p>\n<p>sec A function in terms of sec A:<\/p>\n<p>cos<sup>2<\/sup>A + sin<sup>2<\/sup>A = 1<\/p>\n<p>Rearrange the terms<\/p>\n<p>sin<sup>2<\/sup>A = 1 \u2013 cos<sup>2<\/sup>A<\/p>\n<p>sin<sup>2<\/sup>A = 1 \u2013 (1\/sec<sup>2<\/sup>A)<\/p>\n<p>sin<sup>2<\/sup>A = (sec<sup>2<\/sup>A-1)\/sec<sup>2<\/sup>A<\/p>\n<p>sin A = \u00b1 \u221a(sec<sup>2<\/sup>A-1)\/sec A<\/p>\n<p>cosec A function in terms of sec A:<\/p>\n<p>sin A = 1\/cosec A<\/p>\n<p>\u21d2cosec A = 1\/sin A<\/p>\n<p>cosec A = \u00b1 sec A\/\u221a(sec<sup>2<\/sup>A-1)<\/p>\n<p>Now, tan A function in terms of sec A:<\/p>\n<p>sec<sup>2<\/sup>A \u2013 tan<sup>2<\/sup>A = 1<\/p>\n<p>Rearrange the terms<\/p>\n<p>\u21d2 tan<sup>2<\/sup>A = sec<sup>2<\/sup>A \u2013 1<\/p>\n<p>tan A = \u221a(sec<sup>2<\/sup>A \u2013 1)<\/p>\n<p>cot A function in terms of sec A:<\/p>\n<p>tan A = 1\/cot A<\/p>\n<p>\u21d2 cot A = 1\/tan A<\/p>\n<p>cot A = \u00b11\/\u221a(sec<sup>2<\/sup>A \u2013 1)<\/p>\n<p><strong>3. Evaluate:<\/strong><\/p>\n<p><strong>(i) (sin<sup>2<\/sup>63\u00b0 + sin<sup>2<\/sup>27\u00b0)\/(cos<sup>2<\/sup>17\u00b0 + cos<sup>2<\/sup>73\u00b0)<br \/>\n(ii)  sin 25\u00b0 cos 65\u00b0 + cos 25\u00b0 sin 65\u00b0<\/strong><\/p>\n<p>Solution:<\/p>\n<p>(i) (sin<sup>2<\/sup>63\u00b0 + sin<sup>2<\/sup>27\u00b0)\/(cos<sup>2<\/sup>17\u00b0 + cos<sup>2<\/sup>73\u00b0)<\/p>\n<p>To simplify this, convert some of the sin functions into cos functions and cos function into sin function and it becomes,<\/p>\n<p>= [sin<sup>2<\/sup>(90\u00b0-27\u00b0) + sin<sup>2<\/sup>27\u00b0] \/ [cos<sup>2<\/sup>(90\u00b0-73\u00b0) + cos<sup>2<\/sup>73\u00b0)]\n<p>= (cos<sup>2<\/sup>27\u00b0<sup> <\/sup>+ sin<sup>2<\/sup>27\u00b0)\/(sin<sup>2<\/sup>27\u00b0 + cos<sup>2<\/sup>73\u00b0)<\/p>\n<p>= 1\/1 =1                       (since sin<sup>2<\/sup>A + cos<sup>2<\/sup>A = 1)<\/p>\n<p>Therefore, (sin<sup>2<\/sup>63\u00b0 + sin<sup>2<\/sup>27\u00b0)\/(cos<sup>2<\/sup>17\u00b0 + cos<sup>2<\/sup>73\u00b0) = 1<\/p>\n<p>(ii) sin 25\u00b0 cos 65\u00b0 + cos 25\u00b0 sin 65\u00b0<\/p>\n<p>To simplify this, convert some of the sin functions into cos functions and cos function into sin function and it becomes,<\/p>\n<p>= sin(90\u00b0-25\u00b0) cos 65\u00b0 + cos (90\u00b0-65\u00b0) sin 65\u00b0<\/p>\n<p>= cos 65\u00b0 cos 65\u00b0 + sin 65\u00b0 sin 65\u00b0<\/p>\n<p>= cos<sup>2<\/sup>65\u00b0<sup> <\/sup>+ sin<sup>2<\/sup>65\u00b0 = 1 (since sin<sup>2<\/sup>A + cos<sup>2<\/sup>A = 1)<\/p>\n<p>Therefore, sin 25\u00b0 cos 65\u00b0 + cos 25\u00b0 sin 65\u00b0 = 1<\/p>\n<p><strong>4. Choose the correct option. Justify your choice.<br \/>\n(i) 9 sec<sup>2<\/sup>A \u2013 9 tan<sup>2<\/sup>A =<br \/>\n(A) 1                 (B) 9              (C) 8                (D) 0<br \/>\n(ii) (1 + tan \u03b8 + sec \u03b8) (1 + cot \u03b8 \u2013 cosec \u03b8)<br \/>\n(A) 0                 (B) 1              (C) 2                (D) \u2013 1<br \/>\n(iii) (sec A + tan A) (1 \u2013 sin A) =<br \/>\n(A) sec A           (B) sin A        (C) cosec A      (D) cos A<\/strong><\/p>\n<p><strong>(iv) 1+tan<sup>2<\/sup>A\/1+cot<sup>2<\/sup>A = <\/strong><\/p>\n<p><strong>      (A) sec<sup>2 <\/sup>A                 (B) -1              (C) cot<sup>2<\/sup>A                (D) tan<sup>2<\/sup>A<\/strong><\/p>\n<p>Solution:<\/p>\n<p>(i) (B) is correct.<\/p>\n<p>Justification:<\/p>\n<p>Take 9 outside, and it becomes<\/p>\n<p>9 sec<sup>2<\/sup>A \u2013 9 tan<sup>2<\/sup>A<\/p>\n<p>= 9 (sec<sup>2<\/sup>A \u2013 tan<sup>2<\/sup>A)<\/p>\n<p>= 9\u00d71 = 9             (\u2235 sec2 A \u2013 tan2 A = 1)<\/p>\n<p>Therefore, 9 sec<sup>2<\/sup>A \u2013 9 tan<sup>2<\/sup>A = 9<\/p>\n<p>(ii) (C) is correct<\/p>\n<p>Justification:<\/p>\n<p>(1 + tan \u03b8 + sec \u03b8) (1 + cot \u03b8 \u2013 cosec \u03b8)<\/p>\n<p>We know that, tan \u03b8 = sin \u03b8\/cos \u03b8<\/p>\n<p>sec \u03b8 = 1\/ cos \u03b8<\/p>\n<p>cot \u03b8 = cos \u03b8\/sin \u03b8<\/p>\n<p>cosec \u03b8 = 1\/sin \u03b8<\/p>\n<p>Now, substitute the above values in the given problem, we get<\/p>\n<p>= (1 + sin \u03b8\/cos \u03b8 + 1\/ cos \u03b8) (1 + cos \u03b8\/sin \u03b8 \u2013 1\/sin \u03b8)<\/p>\n<p>Simplify the above equation,<\/p>\n<p>= (cos \u03b8 +sin \u03b8+1)\/cos \u03b8 \u00d7 (sin \u03b8+cos \u03b8-1)\/sin \u03b8<\/p>\n<p>= (cos \u03b8+sin \u03b8)<sup>2<\/sup>-1<sup>2<\/sup>\/(cos \u03b8 sin \u03b8)<\/p>\n<p>= (cos<sup>2<\/sup>\u03b8 + sin<sup>2<\/sup>\u03b8 + 2cos \u03b8 sin \u03b8 -1)\/(cos \u03b8 sin \u03b8)<\/p>\n<p>= (1+ 2cos \u03b8 sin \u03b8 -1)\/(cos \u03b8 sin \u03b8) (Since cos<sup>2<\/sup>\u03b8 + sin<sup>2<\/sup>\u03b8 = 1)<\/p>\n<p>= (2cos \u03b8 sin \u03b8)\/(cos \u03b8 sin \u03b8) = 2<\/p>\n<p>Therefore, (1 + tan \u03b8 + sec \u03b8) (1 + cot \u03b8 \u2013 cosec \u03b8) =2<\/p>\n<p>(iii) (D) is correct.<\/p>\n<p>Justification:<\/p>\n<p>We know that,<\/p>\n<p>Sec A= 1\/cos A<\/p>\n<p>Tan A = sin A \/ cos A<\/p>\n<p>Now, substitute the above values in the given problem, we get<\/p>\n<p>(secA + tanA) (1 \u2013 sinA)<\/p>\n<p>= (1\/cos A + sin A\/cos A) (1 \u2013 sinA)<\/p>\n<p>= (1+sin A\/cos A) (1 \u2013 sinA)<\/p>\n<p>= (1 \u2013 sin<sup>2<\/sup>A)\/cos A<\/p>\n<p>= cos<sup>2<\/sup>A\/cos A = cos A<\/p>\n<p>Therefore, (secA + tanA) (1 \u2013 sinA) = cos A<\/p>\n<p>(iv) (D) is correct.<\/p>\n<p>Justification:<\/p>\n<p>We know that,<\/p>\n<p>tan<sup>2<\/sup>A =1\/cot<sup>2<\/sup>A<\/p>\n<p>Now, substitute this in the given problem, we get<\/p>\n<p>1+tan<sup>2<\/sup>A\/1+cot<sup>2<\/sup>A<\/p>\n<p>= (1+1\/cot<sup>2<\/sup>A)\/1+cot<sup>2<\/sup>A<\/p>\n<p>= (cot<sup>2<\/sup>A+1\/cot<sup>2<\/sup>A)\u00d7(1\/1+cot<sup>2<\/sup>A)<\/p>\n<p>= 1\/cot<sup>2<\/sup>A = tan<sup>2<\/sup>A<\/p>\n<p>So, 1+tan<sup>2<\/sup>A\/1+cot<sup>2<\/sup>A = tan<sup>2<\/sup>A<\/p>\n<p><strong>5. Prove the following identities, where the angles involved are acute angles for which the<br \/>\nexpressions are defined.<\/strong><\/p>\n<p><strong>(i) (cosec \u03b8 \u2013 cot \u03b8)<sup>2 <\/sup>= (1-cos \u03b8)\/(1+cos \u03b8)<\/strong><\/p>\n<p><strong>(ii) cos A\/(1+sin A) + (1+sin A)\/cos A = 2 sec A<\/strong><\/p>\n<p><strong>(iii) tan \u03b8\/(1-cot \u03b8) + cot \u03b8\/(1-tan \u03b8) = 1 + sec \u03b8 cosec \u03b8<\/strong><\/p>\n<p><strong>     [Hint : Write the expression in terms of sin \u03b8 and cos \u03b8]<\/strong><\/p>\n<p><strong>(iv) (1 + sec A)\/sec A = sin<sup>2<\/sup>A\/(1-cos A)  <\/strong><\/p>\n<p><strong>   [Hint : Simplify LHS and RHS separately]<\/strong><\/p>\n<p><strong>(v) ( cos A\u2013sin A+1)\/( cos A +sin A\u20131) = cosec A + cot A, using the identity cosec<sup>2<\/sup>A = 1+cot<sup>2<\/sup>A.<\/strong><\/p>\n<p><img loading=\"lazy\" class=\"alignnone size-full wp-image-711461\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2022\/01\/Screenshot-2024-03-07-130805.png\" alt=\"\" width=\"301\" height=\"75\" \/><\/p>\n<p><strong>(vii) (sin \u03b8 \u2013 2sin<sup>3<\/sup>\u03b8)\/(2cos<sup>3<\/sup>\u03b8-cos \u03b8) = tan \u03b8<br \/>\n(viii) (sin A + cosec A)<sup>2 <\/sup>+ (cos A + sec A)<sup>2<\/sup> = 7+tan<sup>2<\/sup>A+cot<sup>2<\/sup>A<br \/>\n(ix) (cosec A \u2013 sin A)(sec A \u2013 cos A) = 1\/(tan A+cotA)<br \/>\n[Hint : Simplify LHS and RHS separately] (x) (1+tan<sup>2<\/sup>A\/1+cot<sup>2<\/sup>A) = (1-tan A\/1-cot A)<sup>2<\/sup> =<sup> <\/sup>tan<sup>2<\/sup>A<\/strong><\/p>\n<p>Solution:<\/p>\n<p>(i) (cosec \u03b8 \u2013 cot \u03b8)<sup>2 <\/sup>= (1-cos \u03b8)\/(1+cos \u03b8)<\/p>\n<p>To prove this, first take the Left-Hand side (L.H.S) of the given equation, to prove the Right Hand Side (R.H.S)<\/p>\n<p>L.H.S. = (cosec \u03b8 \u2013 cot \u03b8)<sup>2<\/sup><\/p>\n<p>The above equation is in the form of (a-b)<sup>2<\/sup>, and expand it<\/p>\n<p>Since (a-b)<sup>2<\/sup> = a<sup>2<\/sup> + b<sup>2<\/sup> \u2013 2ab<\/p>\n<p>Here a = cosec \u03b8 and b = cot \u03b8<\/p>\n<p>= (cosec<sup>2<\/sup>\u03b8 + cot<sup>2<\/sup>\u03b8 \u2013 2cosec \u03b8 cot \u03b8)<\/p>\n<p>Now, apply the corresponding inverse functions and equivalent ratios to simplify<\/p>\n<p>= (1\/sin<sup>2<\/sup>\u03b8 + cos<sup>2<\/sup>\u03b8\/sin<sup>2<\/sup>\u03b8 \u2013 2cos \u03b8\/sin<sup>2<\/sup>\u03b8)<\/p>\n<p>= (1 + cos<sup>2<\/sup>\u03b8 \u2013 2cos \u03b8)\/(1 \u2013 cos<sup>2<\/sup>\u03b8)<\/p>\n<p>= (1-cos \u03b8)<sup>2<\/sup>\/(1 \u2013 cos\u03b8)(1+cos \u03b8)<\/p>\n<p>= (1-cos \u03b8)\/(1+cos \u03b8) = R.H.S.<\/p>\n<p>Therefore, (cosec \u03b8 \u2013 cot \u03b8)<sup>2 <\/sup>= (1-cos \u03b8)\/(1+cos \u03b8)<\/p>\n<p>Hence proved.<\/p>\n<p>(ii)  (cos A\/(1+sin A)) + ((1+sin A)\/cos A) = 2 sec A<\/p>\n<p>Now, take the L.H.S of the given equation.<\/p>\n<p>L.H.S. = (cos A\/(1+sin A)) + ((1+sin A)\/cos A)<\/p>\n<p>= [cos<sup>2<\/sup>A + (1+sin A)<sup>2<\/sup>]\/(1+sin A)cos A<\/p>\n<p>= (cos<sup>2<\/sup>A + sin<sup>2<\/sup>A + 1 + 2sin A)\/(1+sin A) cos A<\/p>\n<p>Since cos<sup>2<\/sup>A + sin<sup>2<\/sup>A = 1, we can write it as<\/p>\n<p>= (1 + 1 + 2sin A)\/(1+sin A) cos A<\/p>\n<p>= (2+ 2sin A)\/(1+sin A)cos A<\/p>\n<p>= 2(1+sin A)\/(1+sin A)cos A<\/p>\n<p>= 2\/cos A = 2 sec A = R.H.S.<\/p>\n<p>L.H.S. = R.H.S.<\/p>\n<p>(cos A\/(1+sin A)) + ((1+sin A)\/cos A) = 2 sec A<\/p>\n<p>Hence proved.<\/p>\n<p>(iii) tan \u03b8\/(1-cot \u03b8) + cot \u03b8\/(1-tan \u03b8) = 1 + sec \u03b8 cosec \u03b8<\/p>\n<p>L.H.S. = tan \u03b8\/(1-cot \u03b8) + cot \u03b8\/(1-tan \u03b8)<\/p>\n<p>We know that tan \u03b8 =sin \u03b8\/cos \u03b8<\/p>\n<p>cot \u03b8 = cos \u03b8\/sin \u03b8<\/p>\n<p>Now, substitute it in the given equation, to convert it in a simplified form<\/p>\n<p>= [(sin \u03b8\/cos \u03b8)\/1-(cos \u03b8\/sin \u03b8)] + [(cos \u03b8\/sin \u03b8)\/1-(sin \u03b8\/cos \u03b8)]\n<p>= [(sin \u03b8\/cos \u03b8)\/(sin \u03b8-cos \u03b8)\/sin \u03b8] + [(cos \u03b8\/sin \u03b8)\/(cos \u03b8-sin \u03b8)\/cos \u03b8]\n<p>= sin<sup>2<\/sup>\u03b8\/[cos \u03b8(sin \u03b8-cos \u03b8)] + cos<sup>2<\/sup>\u03b8\/[sin \u03b8(cos \u03b8-sin \u03b8)]\n<p>= sin<sup>2<\/sup>\u03b8\/[cos \u03b8(sin \u03b8-cos \u03b8)] \u2013 cos<sup>2<\/sup>\u03b8\/[sin \u03b8(sin \u03b8-cos \u03b8)]\n<p>= 1\/(sin \u03b8-cos \u03b8) [(sin<sup>2<\/sup>\u03b8\/cos \u03b8) \u2013 (cos<sup>2<\/sup>\u03b8\/sin \u03b8)]\n<p>= 1\/(sin \u03b8-cos \u03b8) \u00d7 [(sin<sup>3<\/sup>\u03b8 \u2013 cos<sup>3<\/sup>\u03b8)\/sin \u03b8 cos \u03b8]\n<p>= [(sin \u03b8-cos \u03b8)(sin<sup>2<\/sup>\u03b8+cos<sup>2<\/sup>\u03b8+sin \u03b8 cos \u03b8)]\/[(sin \u03b8-cos \u03b8)sin \u03b8 cos \u03b8]\n<p>= (1 + sin \u03b8 cos \u03b8)\/sin \u03b8 cos \u03b8<\/p>\n<p>= 1\/sin \u03b8 cos \u03b8 + 1<\/p>\n<p>= 1 + sec \u03b8 cosec \u03b8 = R.H.S.<\/p>\n<p>Therefore, L.H.S. = R.H.S.<\/p>\n<p>Hence proved<\/p>\n<p>(iv)  (1 + sec A)\/sec A = sin<sup>2<\/sup>A\/(1-cos A)<\/p>\n<p>First find the simplified form of L.H.S<\/p>\n<p>L.H.S. = (1 + sec A)\/sec A<\/p>\n<p>Since secant function is the inverse function of cos function and it is written as<\/p>\n<p>= (1 + 1\/cos A)\/1\/cos A<\/p>\n<p>= (cos A + 1)\/cos A\/1\/cos A<\/p>\n<p>Therefore, (1 + sec A)\/sec A = cos A + 1<\/p>\n<p>R.H.S. = sin<sup>2<\/sup>A\/(1-cos A)<\/p>\n<p>We know that sin<sup>2<\/sup>A = (1 \u2013 cos<sup>2<\/sup>A), we get<\/p>\n<p>= (1 \u2013 cos<sup>2<\/sup>A)\/(1-cos A)<\/p>\n<p>= (1-cos A)(1+cos A)\/(1-cos A)<\/p>\n<p>Therefore, sin<sup>2<\/sup>A\/(1-cos A)= cos A + 1<\/p>\n<p>L.H.S. = R.H.S.<\/p>\n<p>Hence proved<\/p>\n<p>(v) (cos A\u2013sin A+1)\/(cos A+sin A\u20131) = cosec A + cot A, using the identity cosec<sup>2<\/sup>A = 1+cot<sup>2<\/sup>A.<\/p>\n<p>With the help of identity function, cosec<sup>2<\/sup>A = 1+cot<sup>2<\/sup>A, let us prove the above equation.<\/p>\n<p>L.H.S. = (cos A\u2013sin A+1)\/(cos A+sin A\u20131)<\/p>\n<p>Divide the numerator and denominator by sin A, we get<\/p>\n<p>= (cos A\u2013sin A+1)\/sin A\/(cos A+sin A\u20131)\/sin A<\/p>\n<p>We know that cos A\/sin A = cot A and 1\/sin A = cosec A<\/p>\n<p>= (cot A \u2013 1 + cosec A)\/(cot A+ 1 \u2013 cosec A)<\/p>\n<p>= (cot A \u2013 cosec<sup>2<\/sup>A + cot<sup>2<\/sup>A + cosec A)\/(cot A+ 1 \u2013 cosec A) (using cosec<sup>2<\/sup>A \u2013 cot<sup>2<\/sup>A = 1<\/p>\n<p>= [(cot A + cosec A) \u2013 (cosec<sup>2<\/sup>A \u2013 cot<sup>2<\/sup>A)]\/(cot A+ 1 \u2013 cosec A)<\/p>\n<p>= [(cot A + cosec A) \u2013 (cosec A + cot A)(cosec A \u2013 cot A)]\/(1 \u2013 cosec A + cot A)<\/p>\n<p>=  (cot A + cosec A)(1 \u2013 cosec A + cot A)\/(1 \u2013 cosec A + cot A)<\/p>\n<p>=  cot A + cosec A = R.H.S.<\/p>\n<p>Therefore, (cos A\u2013sin A+1)\/(cos A+sin A\u20131) = cosec A + cot A<\/p>\n<p>Hence Proved<\/p>\n<p><img loading=\"lazy\" class=\"alignnone wp-image-711462\" src=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2022\/01\/Screenshot-2024-03-07-130909-300x275.png\" alt=\"\" width=\"364\" height=\"334\" srcset=\"https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2022\/01\/Screenshot-2024-03-07-130909-300x275.png 300w, https:\/\/infinitylearn.com\/surge\/wp-content\/uploads\/2022\/01\/Screenshot-2024-03-07-130909.png 760w\" sizes=\"(max-width: 364px) 100vw, 364px\" \/><\/p>\n<p>= (sec A + tan A)\/1<\/p>\n<p>= sec A + tan A = R.H.S<\/p>\n<p>Hence proved<\/p>\n<p>(vii) (sin \u03b8 \u2013 2sin<sup>3<\/sup>\u03b8)\/(2cos<sup>3<\/sup>\u03b8-cos \u03b8) = tan \u03b8<\/p>\n<p>L.H.S. = (sin \u03b8 \u2013 2sin<sup>3<\/sup>\u03b8)\/(2cos<sup>3<\/sup>\u03b8 \u2013 cos \u03b8)<\/p>\n<p>Take sin \u03b8 as in numerator and cos \u03b8 in denominator as outside, it becomes<\/p>\n<p>= [sin \u03b8(1 \u2013 2sin<sup>2<\/sup>\u03b8)]\/[cos \u03b8(2cos<sup>2<\/sup>\u03b8- 1)]\n<p>We know that sin<sup>2<\/sup>\u03b8 = 1-cos<sup>2<\/sup>\u03b8<\/p>\n<p>= sin \u03b8[1 \u2013 2(1-cos<sup>2<\/sup>\u03b8)]\/[cos \u03b8(2cos<sup>2<\/sup>\u03b8 -1)]\n<p>= [sin \u03b8(2cos<sup>2<\/sup>\u03b8 -1)]\/[cos \u03b8(2cos<sup>2<\/sup>\u03b8 -1)]\n<p>= tan \u03b8 = R.H.S.<\/p>\n<p>Hence proved<\/p>\n<p>(viii) (sin A + cosec A)<sup>2 <\/sup>+ (cos A + sec A)<sup>2<\/sup> = 7+tan<sup>2<\/sup>A+cot<sup>2<\/sup>A<\/p>\n<p>L.H.S. = (sin A + cosec A)<sup>2 <\/sup>+ (cos A + sec A)<sup>2<\/sup><\/p>\n<p>It is of the form (a+b)<sup>2<\/sup>, expand it<\/p>\n<p>(a+b)<sup>2<\/sup> =a<sup>2<\/sup> + b<sup>2<\/sup> +2ab<\/p>\n<p>= (sin<sup>2<\/sup>A + cosec<sup>2<\/sup>A + 2 sin A cosec A) + (cos<sup>2<\/sup>A + sec<sup>2<\/sup>A + 2 cos A sec A)<\/p>\n<p>= (sin<sup>2<\/sup>A + cos<sup>2<\/sup>A) + 2 sin A(1\/sin A) + 2 cos A(1\/cos A) + 1 + tan<sup>2<\/sup>A + 1 + cot<sup>2<\/sup>A<\/p>\n<p>= 1 + 2 + 2 + 2 + tan<sup>2<\/sup>A + cot<sup>2<\/sup>A<\/p>\n<p>= 7+tan<sup>2<\/sup>A+cot<sup>2<\/sup>A = R.H.S.<\/p>\n<p>Therefore, (sin A + cosec A)<sup>2 <\/sup>+ (cos A + sec A)<sup>2<\/sup> = 7+tan<sup>2<\/sup>A+cot<sup>2<\/sup>A<\/p>\n<p>Hence proved.<\/p>\n<p>(ix) (cosec A \u2013 sin A)(sec A \u2013 cos A) = 1\/(tan A+cotA)<\/p>\n<p>First, find the simplified form of L.H.S<\/p>\n<p>L.H.S. = (cosec A \u2013 sin A)(sec A \u2013 cos A)<\/p>\n<p>Now, substitute the inverse and equivalent trigonometric ratio forms<\/p>\n<p>= (1\/sin A \u2013 sin A)(1\/cos A \u2013 cos A)<\/p>\n<p>= [(1-sin<sup>2<\/sup>A)\/sin A][(1-cos<sup>2<\/sup>A)\/cos A]\n<p>= (cos<sup>2<\/sup>A\/sin A)\u00d7(sin<sup>2<\/sup>A\/cos A)<\/p>\n<p>= cos A sin A<\/p>\n<p>Now, simplify the R.H.S<\/p>\n<p>R.H.S. = 1\/(tan A+cotA)<\/p>\n<p>= 1\/(sin A\/cos A +cos A\/sin A)<\/p>\n<p>= 1\/[(sin<sup>2<\/sup>A+cos<sup>2<\/sup>A)\/sin A cos A]\n<p>= cos A sin A<\/p>\n<p>L.H.S. = R.H.S.<\/p>\n<p>(cosec A \u2013 sin A)(sec A \u2013 cos A) = 1\/(tan A+cotA)<\/p>\n<p>Hence proved<\/p>\n<p>(x)  (1+tan<sup>2<\/sup>A\/1+cot<sup>2<\/sup>A) = (1-tan A\/1-cot A)<sup>2<\/sup> =<sup> <\/sup>tan<sup>2<\/sup>A<\/p>\n<p>L.H.S. = (1+tan<sup>2<\/sup>A\/1+cot<sup>2<\/sup>A)<\/p>\n<p>Since cot function is the inverse of tan function,<\/p>\n<p>= (1+tan<sup>2<\/sup>A\/1+1\/tan<sup>2<\/sup>A)<\/p>\n<p>= 1+tan<sup>2<\/sup>A\/[(1+tan<sup>2<\/sup>A)\/tan<sup>2<\/sup>A]\n<p>Now cancel the 1+tan<sup>2<\/sup>A terms, we get<\/p>\n<p>= tan<sup>2<\/sup>A<\/p>\n<p>(1+tan<sup>2<\/sup>A\/1+cot<sup>2<\/sup>A) = tan<sup>2<\/sup>A<\/p>\n<p>Similarly,<\/p>\n<p>(1-tan A\/1-cot A)<sup>2<\/sup> =<sup> <\/sup>tan<sup>2<\/sup>A<\/p>\n<p>Hence proved<\/p>\n<h2><span class=\"ez-toc-section\" id=\"FAQs_on_Introduction_to_Trigonometry_-_Class_10_Maths_Chapter_8\"><\/span><strong>FAQs on Introduction to Trigonometry &#8211; Class 10 Maths Chapter 8<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n\t\t<section class=\"sc_fs_faq sc_card \">\n\t\t\t<div>\n\t\t\t\t<h3><span class=\"ez-toc-section\" id=\"What_is_Trigonometry_in_Class_10_Maths\"><\/span>What is Trigonometry in Class 10 Maths?<span class=\"ez-toc-section-end\"><\/span><\/h3>\t\t\t\t<div>\n\t\t\t\t\t\t\t\t\t\t<p>\n\t\t\t\t\t\tTrigonometry is a branch of mathematics that deals with the relationships between angles and sides of triangles.\t\t\t\t\t<\/p>\n\t\t\t\t<\/div>\n\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"sc_fs_faq sc_card \">\n\t\t\t<div>\n\t\t\t\t<h3><span class=\"ez-toc-section\" id=\"Why_is_Trigonometry_important_in_Class_10_Maths\"><\/span>Why is Trigonometry important in Class 10 Maths?<span class=\"ez-toc-section-end\"><\/span><\/h3>\t\t\t\t<div>\n\t\t\t\t\t\t\t\t\t\t<p>\n\t\t\t\t\t\tTrigonometry helps solve real-world problems involving measurements of angles and sides, making it a practical skill.\t\t\t\t\t<\/p>\n\t\t\t\t<\/div>\n\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"sc_fs_faq sc_card \">\n\t\t\t<div>\n\t\t\t\t<h3><span class=\"ez-toc-section\" id=\"What_concepts_are_covered_in_Class_10_Trigonometry\"><\/span>What concepts are covered in Class 10 Trigonometry?<span class=\"ez-toc-section-end\"><\/span><\/h3>\t\t\t\t<div>\n\t\t\t\t\t\t\t\t\t\t<p>\n\t\t\t\t\t\tClass 10 Trigonometry covers topics like trigonometric ratios, Pythagoras theorem, and applications of trigonometry.\t\t\t\t\t<\/p>\n\t\t\t\t<\/div>\n\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"sc_fs_faq sc_card \">\n\t\t\t<div>\n\t\t\t\t<h3><span class=\"ez-toc-section\" id=\"What_are_Trigonometric_Ratios\"><\/span>What are Trigonometric Ratios?<span class=\"ez-toc-section-end\"><\/span><\/h3>\t\t\t\t<div>\n\t\t\t\t\t\t\t\t\t\t<p>\n\t\t\t\t\t\tTrigonometric ratios relate angles to sides of a right triangle. They include sine, cosine, and tangent.\t\t\t\t\t<\/p>\n\t\t\t\t<\/div>\n\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"sc_fs_faq sc_card \">\n\t\t\t<div>\n\t\t\t\t<h3><span class=\"ez-toc-section\" id=\"How_can_Trigonometry_be_applied_in_real_life\"><\/span>How can Trigonometry be applied in real life?<span class=\"ez-toc-section-end\"><\/span><\/h3>\t\t\t\t<div>\n\t\t\t\t\t\t\t\t\t\t<p>\n\t\t\t\t\t\tTrigonometry is used in fields like architecture, engineering, astronomy, and navigation for calculations involving angles and distances.\t\t\t\t\t<\/p>\n\t\t\t\t<\/div>\n\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"sc_fs_faq sc_card \">\n\t\t\t<div>\n\t\t\t\t<h3><span class=\"ez-toc-section\" id=\"What_is_Pythagoras_Theorem_in_Trigonometry\"><\/span>What is Pythagoras Theorem in Trigonometry?<span class=\"ez-toc-section-end\"><\/span><\/h3>\t\t\t\t<div>\n\t\t\t\t\t\t\t\t\t\t<p>\n\t\t\t\t\t\tPythagoras Theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.\t\t\t\t\t<\/p>\n\t\t\t\t<\/div>\n\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"sc_fs_faq sc_card \">\n\t\t\t<div>\n\t\t\t\t<h3><span class=\"ez-toc-section\" id=\"How_are_Trigonometric_Ratios_calculated\"><\/span>How are Trigonometric Ratios calculated?<span class=\"ez-toc-section-end\"><\/span><\/h3>\t\t\t\t<div>\n\t\t\t\t\t\t\t\t\t\t<p>\n\t\t\t\t\t\tTrigonometric ratios are calculated by dividing the lengths of specific sides of a right triangle.\t\t\t\t\t<\/p>\n\t\t\t\t<\/div>\n\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"sc_fs_faq sc_card \">\n\t\t\t<div>\n\t\t\t\t<h3><span class=\"ez-toc-section\" id=\"What_are_the_primary_applications_of_Trigonometry\"><\/span>What are the primary applications of Trigonometry?<span class=\"ez-toc-section-end\"><\/span><\/h3>\t\t\t\t<div>\n\t\t\t\t\t\t\t\t\t\t<p>\n\t\t\t\t\t\tTrigonometry is used for calculating heights, distances, angles of elevation, and depression in real-life scenarios.\t\t\t\t\t<\/p>\n\t\t\t\t<\/div>\n\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"sc_fs_faq sc_card \">\n\t\t\t<div>\n\t\t\t\t<h3><span class=\"ez-toc-section\" id=\"How_can_I_understand_Trigonometry_concepts_better\"><\/span>How can I understand Trigonometry concepts better?<span class=\"ez-toc-section-end\"><\/span><\/h3>\t\t\t\t<div>\n\t\t\t\t\t\t\t\t\t\t<p>\n\t\t\t\t\t\tPractice with examples, diagrams, and real-life scenarios helps in better understanding Trigonometry concepts.\t\t\t\t\t<\/p>\n\t\t\t\t<\/div>\n\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<section class=\"sc_fs_faq sc_card \">\n\t\t\t<div>\n\t\t\t\t<h3><span class=\"ez-toc-section\" id=\"Is_Trigonometry_covered_in_other_math_classes_too\"><\/span>Is Trigonometry covered in other math classes too?<span class=\"ez-toc-section-end\"><\/span><\/h3>\t\t\t\t<div>\n\t\t\t\t\t\t\t\t\t\t<p>\n\t\t\t\t\t\tYes, Trigonometry concepts are introduced in higher math classes and continue to be applied in advanced mathematics.\t\t\t\t\t<\/p>\n\t\t\t\t<\/div>\n\t\t\t<\/div>\n\t\t<\/section>\n\t\t\n<script type=\"application\/ld+json\">\n\t{\n\t\t\"@context\": \"https:\/\/schema.org\",\n\t\t\"@type\": \"FAQPage\",\n\t\t\"mainEntity\": [\n\t\t\t\t\t{\n\t\t\t\t\"@type\": \"Question\",\n\t\t\t\t\"name\": \"What is Trigonometry in Class 10 Maths?\",\n\t\t\t\t\"acceptedAnswer\": {\n\t\t\t\t\t\"@type\": \"Answer\",\n\t\t\t\t\t\"text\": \"Trigonometry is a branch of mathematics that deals with the relationships between angles and sides of triangles.\"\n\t\t\t\t\t\t\t\t\t}\n\t\t\t}\n\t\t\t,\t\t\t\t{\n\t\t\t\t\"@type\": \"Question\",\n\t\t\t\t\"name\": \"Why is Trigonometry important in Class 10 Maths?\",\n\t\t\t\t\"acceptedAnswer\": {\n\t\t\t\t\t\"@type\": \"Answer\",\n\t\t\t\t\t\"text\": \"Trigonometry helps solve real-world problems involving measurements of angles and sides, making it a practical skill.\"\n\t\t\t\t\t\t\t\t\t}\n\t\t\t}\n\t\t\t,\t\t\t\t{\n\t\t\t\t\"@type\": \"Question\",\n\t\t\t\t\"name\": \"What concepts are covered in Class 10 Trigonometry?\",\n\t\t\t\t\"acceptedAnswer\": {\n\t\t\t\t\t\"@type\": \"Answer\",\n\t\t\t\t\t\"text\": \"Class 10 Trigonometry covers topics like trigonometric ratios, Pythagoras theorem, and applications of trigonometry.\"\n\t\t\t\t\t\t\t\t\t}\n\t\t\t}\n\t\t\t,\t\t\t\t{\n\t\t\t\t\"@type\": \"Question\",\n\t\t\t\t\"name\": \"What are Trigonometric Ratios?\",\n\t\t\t\t\"acceptedAnswer\": {\n\t\t\t\t\t\"@type\": \"Answer\",\n\t\t\t\t\t\"text\": \"Trigonometric ratios relate angles to sides of a right triangle. 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Download a free pdf of the chapter on Infinity Learn.","custom_permalink":"study-materials\/maths\/introduction-to-trigonometry\/class-10-extra-questions\/maths-chapter-8\/"},"categories":[13,21],"tags":[],"table_tags":[],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v17.9 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry<\/title>\n<meta name=\"description\" content=\"Here is an Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8. 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