Important Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry

# Important Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry

## Important Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry

### Introduction to Trigonometry Class 10 Important Questions Very Short Answer (1 Mark)

Question 1.
If tan θ + cot θ = 5, find the value of tan2θ + cotθ. (2012)
Solution:
tan θ + cot θ = 5 … [Given
tan2θ + cot2θ + 2 tan θ cot θ = 25 … [Squaring both sides
tan2θ + cot2θ + 2 = 25
∴ tan2θ + cot2θ = 23

Fill Out the Form for Expert Academic Guidance!

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)

Question 2.
If sec 2A = cosec (A – 27°) where 2A is an acute angle, find the measure of ∠A. (2012, 2017D)
Solution:
sec 2A = cosec (A – 27°)
cosec(90° – 2A) = cosec(A – 27°) …[∵ sec θ = cosec (90° – θ)
90° – 2A = A – 27°
90° + 27° = 2A + A
⇒ 3A = 117°
∴ ∠A = $$\frac{117^{\circ}}{3}$$ = 39°

Question 3.
If tan α = $$\sqrt{3}$$ and tan β = $$\frac{1}{\sqrt{3}}$$,0 < α, β < 90°, find the value of cot (α + β). (2012)
Solution:
tan α = $$\sqrt{3}$$ = tan 60° …(i)
tan β = $$\frac{1}{\sqrt{3}}$$ = tan 30° …(ii)
Solving (i) & (ii), α = 60° and β = 30°
∴ cot (α + β) = cot (60° + 30°) = cot 90° = 0

Question 4.
If sin θ – cos θ = 0, find the value of sin4 θ + cos4 θ. (2012, 2017D)
Solution:
sin θ – cos θ = 0 = sin θ = cos θ
⇒ $$\frac{\sin \theta}{\cos \theta}$$ = 1 ⇒ tan θ = 1 ⇒ θ = 45°
Now, sin4θ + cos4θ
= sin4 45° + cos4 45°
= $$\left(\frac{1}{\sqrt{2}}\right)^{4}+\left(\frac{1}{\sqrt{2}}\right)^{4}=\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$$

Question 5.
If sec θ + tan θ = 7, then evaluate sec θ – tan θ. (2017OD)
Solution:
We know that,
sec2θ – tan2θ = 1
(sec θ + tan θ) (sec θ – tan θ) = 1
(7) (sec θ – tan θ) = 1 …[sec θ + tan θ = 7; (Given)
∴ sec θ – tan θ = $$\frac{1}{7}$$

Question 6.
Evaluate: 10. $$\frac{1-\cot ^{2} 45^{\circ}}{1+\sin ^{2} 90^{\circ}}$$. (2014)
Solution:

Question 7.
If cosec θ = $$\frac{5}{4}$$, find the value of cot θ. (2014)
Solution:
We know that, cot2θ = cosec2θ – 1
= $$\left(\frac{5}{4}\right)^{2}$$ – 1 ⇒ $$\frac{25}{16}$$ – 1 ⇒ $$\frac{25-16}{16}$$
coť2θ = $$\frac{9}{16}$$ i cot θ = $$\frac{3}{4}$$

Question 8.
If θ = 45°, then what is the value of 2 sec2θ + 3 cosec2θ ? (2014)
Solution:
2 sec2θ + 3 cosec2θ = 2 sec2 45° + 3 cosec2 45°
= 2($$\sqrt{2}$$)2 + 3 ($$\sqrt{2}$$)2 = 4 + 6 = 10

Question 9.
If $$\sqrt{3}$$ sin θ = cos θ, find the value of $$\frac{3 \cos ^{2} \theta+2 \cos \theta}{3 \cos \theta+2}$$. (2015)
Solution:
$$\sqrt{3}$$ sin θ = cos θ … [Given

Question 10.
Evaluate: sin2 19° + sin771°. (2015)
Solution:
sin2 19° + sin2 71°
= sin219° + sin2 (90° – 19°)…[∵ sin(90° – θ) = cos θ
= sin2 19° + cos2 19° = 1 …[∵ sin2 θ + cos2 θ = 1

Question 11.
What happens to value of cos when increases from 0° to 90°? (2015)
Solution:
cos 0° = 1, cos 90° = 0
When θ increases from 0° to 90°, the value of cos θ decreases from 1 to 0.

Question 12.
If tan θ = $$\frac{a}{x}$$, find the value of $$\frac{x}{\sqrt{a^{2}+x^{2}}}$$. (2013)
Solution:

Question 13.
If in a right angled ∆ABC, tan B = $$\frac{12}{5}$$, then find sin B. (2014)
Solution:
1st method:
tan B = $$\frac{12}{5}$$ ∴ cot B = $$\frac{5}{12}$$
cosec2 B = 1 + cot2 B
= 1 + $$[\left(\frac{5}{12}\right)^{2}/latex] = 1 + $$
= $$\frac{144+25}{144}=\frac{169}{144}$$
cosec B = $$\frac{13}{12}$$ ∴ sin B = $$\frac{12}{13}$$
2nd method:

tan B = $$\frac{12}{5}$$
tan B = $$\frac{AC}{BC}$$
Let AC = 12k, BC = 5k
In rt. ∆ACB,
AB2 = AC2 + BC2 …[Pythagoras theorem
AB2 = (12k)2 + (5k)2
AB2 = 144k2 + 25k22 = 169k2
AB = 13k
∴ sin B = $$\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{12 \mathrm{k}}{13 \mathrm{k}}=\frac{12}{13}$$

Question 14.
If ∆ABC is right angled at B, what is the value of sin (A + C). (2015)
Solution:

∠B = 90° …[Given
∠A + ∠B + ∠C = 180° …[Angle sum property of a ∆
∠A + ∠C + 90° = 180°
∠A + ∠C = 90°
∴ sin (A + C) = sin 90° = 1 …(taking sin both side

### Introduction to Trigonometry Class 10 Important Questions Short Answer-I (2 Marks)

Question 15.
Evaluate: tan 15° . tan 25° , tan 60° . tan 65° . tan 75° – tan 30°. (2013)
Solution:
tan 15°. tan 25°, tan 60°. tan 65°. tan 75° – tan 30°
= tan(90° – 75°) tan(90° – 65°). $$\sqrt{3}$$ . tan 65°. tan 75° – $$\frac{1}{\sqrt{3}}$$

Question 16.
Express cot 75° + cosec 75° in terms of trigonometric ratios of angles between 0° and 30°. (2013)
Solution:
cot 75° + cosec 75°
= cot(90° – 15°) + cosec(90° – 15°)
= tan 15° + sec 15° …[cot(90°-A) = tan A
cosec(90° – A) = sec A

Question 17.
If cos (A + B) = 0 and sin (A – B) = 3, then find the value of A and B where A and B are acute angles. (2012)
Solution:

Putting the value of B in (i), we get
⇒ A = 30° + 30° = 60°
∴ A = 60°, B = 30°

Question 18.
If A, B and C are the interior angles of a ∆ABC, show that sin $$\left(\frac{A+B}{2}\right)$$ = cos$$\left(\frac{c}{2}\right)$$. (2012)
Solution:
In ∆ABC, ∠A + ∠B + ∠C = 180° …(Angle sum property of ∆
∠A + ∠B = 180° – ∠C

Question 19.
If x = p sec θ + q tan θ and y = p tan θ + q sec θ, then prove that x2 – y2 = p2 – q2. (2014)
Solution:
L.H.S. = x2 – y2
= (p sec θ + q tan θ)2 – (p tan θ + q sec θ)2
= p2 sec θ + q2 tan2 θ + 2 pq sec 2 tan 2 -(p2 tan2 θ + q2 sec2 θ + 2pq sec θ tan θ)
= p2 sec θ + 2 tan2 θ + 2pq sec θ tan θ – p2 tan2 θ – q2 sec θ – 2pq sec θ tan θ
= p2(sec2 θ – tan2 θ) – q2(sec?2 θ – tan2 θ) =
= p2 – q2 …[sec2 θ – tan2 θ = 1
= R.H.S.

Question 20.
Prove the following identity: (2015)
$$\frac{\sin ^{3} \theta+\cos ^{3} \theta}{\sin \theta+\cos \theta}$$ = 1 – sin θ . cos θ
Solution:

Question 21.
Simplify: $$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}$$. (2014)
Solution:

Question 22.
If x = a cos θ – b sin θ and y = a sin θ + b cos θ, then prove that a2 + b2 = x2 + y2. (2015)
Solution:
R.H.S. = x2 + y2
= (a cos θ – b sin θ)2 + (a sin θ + b cos θ)2
= a2cos2 θ + b2 sin2 θ – 2ab cos θ sin θ + a2 sin2 θ + b2 cos2 θ + 2ab sin θ cos θ
= a2(cos2 θ + sin2θ) + b2 (sin2 θ + cos2 θ)
= a2 + b2 = L.H.S. …[∵ cos2 θ + sin2 θ = 1

### Introduction to Trigonometry Class 10 Important Questions Short Answer – II (3 Marks)

Question 23.
Given 2 cos 3θ = $$\sqrt{3}$$, find the value of θ. (2014)
Solution:
2 cos 3θ = $$\sqrt{3}$$ …[Given
cos 3θ = $$\frac{\sqrt{3}}{2}$$ ⇒ cos 3θ = cos 30°
30 = 30° ∴ θ = 10°

Question 24.
If cos x = cos 40° . sin 50° + sin 40°. cos 50°, then find the value of x. (2014)
Solution:
cos x = cos 40° sin 50° + sin 40° cos 50°
cos x = cos 40° sin(90° – 40°) + sin 40°.cos(90° – 40°)
cos x = cos2 40° + sin2 40°
cos x = 1 …[∵ cos2 A + sin2 A = 1
cos x = cos 0° ⇒ x = 0°

Question 25.
If sin θ = $$\frac{1}{2}$$, then show that 3 cos θ – 4 cos3 θ = 0. (2014)
Solution:
sin θ = $$\frac{1}{2}$$
sin θ = sin 30° ⇒ θ = 30°
L.H.S = 3 cos θ – 4 cos3 θ
= 3 cos 30° – 4 cos3(30°)

Question 26.
If 5 sin θ = 4, prove that $$\frac{1}{\cos \theta}+\frac{1}{\cot \theta}$$ = 3 (2013
Solution:
Given: 5 sin θ = 4

Question 27.
Evaluate: sec 41°. sin 49° + cos 29°.cosec 61° (2012)
Solution:

Question 28.
Evaluate: (2012, 2017D)

Solution:

Question 29.
In figure, ∆PQR right angled at Q, PQ = 6 cm and PR = 12 cm. Determine ∠QPR and ∠PRQ. (2013)

Solution:
In rt. ∆PQR,
PQ2 + QR2 = PR2 …[By Pythogoras’ theorem

(6)2 + QR2 = (12)2
QR2 = 144 – 36
QR2 = 108

Question 30.
Find the value of: (2013)

Solution:

Question 31.
Prove that: $$\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\sec ^{2} 20^{\circ}-\cot ^{2} 70^{\circ}}$$ + 2 sin 36° sin 42° sec 48° sec 54° (2017OD)
Solution:

Question 32.
If sin θ = $$\frac{12}{13}$$, 0° <0 < 90°, find the value of: $$\frac{\sin ^{2} \theta-\cos ^{2} \theta}{2 \sin \theta \cdot \cos \theta} \times \frac{1}{\tan ^{2} \theta}$$ (2015)
Solution:

Question 33.
Prove that: (2012)

Solution:

Question 34.
Prove that: $$\frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1}=\frac{1+\sin \theta}{\cos \theta}$$ (2012, 2017D)
Solution:

Question 35.
If tan θ = $$\frac{a}{b}$$, prove that $$\frac{a \sin \theta-b \cos \theta}{a \sin \theta+b \cos \theta}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$$ (2013)
Solution:

Question 36.
Prove the identity: (sec A – cos A). (cot A + tan A) = tan A . sec A. (2014)
Solution:
L.H.S.= (sec A – cos A) (cot A + tan A)

Question 37.
If sec θ + tan θ = p, prove that sin θ = $$\frac{p^{2}-1}{p^{2}+1}$$ (2015)
Solution:

Question 38.
Prove that: $$\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}$$ = tan θ (2015)
Solution:

Question 39.
Prove that: $$\frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}$$ = 2 cosec θ (2017OD)
Solution:

### Introduction to Trigonometry Class 10 Important Questions Long Answer (4 Marks)

Question 40.
In an acute angled triangle ABC, if sin (A + B – C) = $$\frac{1}{2}$$ and cos (B + C – A) = $$\frac{1}{\sqrt{2}}$$, find ∠A, ∠B and ∠C. (2012)
Solution:

Putting the values of A and B in (iii), we get
67.5° + B + 75o = 180°
B = 180° – 67.5° – 75o = 37.5°
∴ ∠A = 67.5°, ∠B = 37.5° and ∠C = 75°

Question 41.
Evaluate: (2013)

Solution:

Question 42.
Evaluate the following: (2015)

Solution:

Question 43.
If θ = 30°, verify the following: (2014)
(i) cos 3θ = 4 cos3 θ – 3 cos θ
(ii) sin 3θ = 3 sin θ – 4 sin3θ
Solution:

Question 44.
If tan (A + B) = $$\sqrt{3}$$ and tan (A – B) = $$\frac{1}{\sqrt{3}}$$ where 0 < A + B < 90°, A > B, find A and B. Also calculate: tan A. sin (A + B) + cos A. tan (A – B). (2015)
Solution:

Question 45.
Find the value of cos 60° geometrically. Hence find cosec 60°. (2012, 2017D)
Solution:

Let ∆ABC be an equilateral ∆.
Let each side of triangle be 2a.
Since each angle in an equilateral ∆ is 60°
∴ ∠A = ∠B = ∠C = 60°
AB = AC … [Each = 2a
∠1 -∠2 … [Each 90°
BD = DC = $$\frac{2 a}{2}$$ = a
In rt. ∆ADB, cos 60° = $$\frac{\mathrm{BD}}{\mathrm{AB}}=\frac{a}{2 a}=\frac{1}{2}$$

Question 46.
If tan(20° – 3α) = cot(5α – 20°), then find the value of α and hence evaluate: sin α. sec α . tan α – cosec α . cos α . cot α. (2014)
Solution:
tan(20° – 3α) = cot(5α – 20°)
tan(20° – 3α) = tan[90° – (5α – 20°)] …[∵ cot θ = tan(90° – θ)] ∴ 20° – 3α = 90° – 5α + 20°
⇒ -3α + 5α = 90° + 20° – 20°
⇒ 2α = 90° ⇒ α = 45°
Now, sin α . sec α tan α – cosec α . cos α . cot α
= sin 45°. sec 45° tan 45° – cosec 45°. cos 45° cot 45°
= $$\frac{1}{\sqrt{2}} \times \sqrt{2} \times 1-\sqrt{2} \times \frac{1}{\sqrt{2}} \times 1=1-1=0$$

Question 47.
If $$\frac{x}{a}$$cosθ + $$\frac{y}{b}$$sinθ = 1 and $$\frac{x}{a}$$sinθ – $$\frac{y}{b}$$ cosθ = 1, prove that event $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$$ = 2. (2012, 2017D)
Solution:

Question 48.
If sin θ = $$\frac{c}{\sqrt{c^{2}+d^{2}}}$$ and d > 0, find the values of cos θ and tan θ. (2013)
Solution:

Question 49.
If cot B = $$\frac{12}{5}$$, prove that tan2B – sin2B = sin4 B . sec2 B. (2013)
Solution:
cot B = $$\frac{12}{5}$$ :: $$\frac{A B}{B C}=\frac{12}{5}$$

AB = 12k, BC = 5k
In rt. ∆ABC, …[By Pythagoras’ theorem
AC2 = AB2 + BC2
AC2 = (12k)2 + (5k)2
AC2 = 144k2 + 25k2
AC2 = 169k2
AC = +13k …[∵ Hypotenuse cannot be -ve

Question 50.
If $$\sqrt{3}$$ cot2θ – 4 cot θ + $$\sqrt{3}$$ = 0, then find the value of cot2 θ + tan2θ. (2013)
Solution:

Question 51.
Prove that b2x2 – a2y2 = a2b2, if: (2014)
(i) x = a sec θ, y = b tan θ
(ii) x = a cosec θ, y = b cot θ
Solution:
(i) L.H.S. = b2x2 – a2y2
= b2(a sec θ)2 – a2(b tan θ)2
= b2a2 sec θ – a2b2 tan2θ
= b2a2(sec2 θ – tan2 θ)
= b2a2(1) …[∵ sec2θ – tan2 θ = 1
= a2b2 = R.H.S.

(ii) L.H.S. = b2x2 – a2y2
= b2(a cosec θ)2 – a2(b cot θ)2
= b2a2 cosec2 θ – a2b2 cot2 θ
= b2a2(cosec2θ – cot2 θ)
= b2a2 (1) ..[∵ cosec2 θ – cot2 θ = 1
= a2b2= R.H.S.

Question 52.
If sec θ – tan θ = x, show that sec θ + tan θ = $$\frac{1}{x}$$ and hence find the values of cos θ and sin θ. (2015)
Solution:

Question 53.
If cosec θ + cot θ = p, then prove that cos θ = $$\frac{p^{2}-1}{p^{2}+1}$$. (2012)
Solution:
cosec θ + cot θ = p

Question 54.
If tan θ + sin θ = p; tan θ – sin θ = q; prove that p2 – q2 = $$4 \sqrt{p q}$$. (2012)
Solution:
L.H.S. = p2 – q2
= (tan θ + sin θ)2 – (tan θ – sin θ)2
= (tan2θ + sin2θ + 2.tanθ.sinθ) – (tan2θ + sin2θ – 2tan θ sin θ)
= 2 tan θ sin θ+ 2 tan θ sin θ
= 4 tan θ sin θ …(i)

Question 55.
If sin θ + cos θ = m and sec θ + cosec θ = n, then prove that n(m2 – 1) = 2m. (2013)
Solution:
m2 – 1 = (sin θ + cos θ)2 – 1
= sin2 θ + cos2θ + 2 sin θ cos θ – 1
= 1 + 2 sin θ cos θ – 1
= 2 sin θ cos θ …[sin2 θ + cos2 θ = 1
L.H.S. = n(m2 – 1)
= (sec θ + cosec θ) 2 sin θ cos θ

Question 56.
Prove that: = 2 cosec A (2012)
Solution:

Question 57.
In ∆ABC, show that sin2 $$\frac{\mathbf{A}}{2}$$ + sin2 $$\frac{\mathbf{B}+\mathbf{C}}{\mathbf{2}}$$ = 1. (2013)
Solution:
In ∆ABC, ∠A + ∠B + ∠C = 180° … [Sum of the angles of ∆
∠B + ∠C = 180° – ∠A

Question 58.
Find the value of: (2013)

Solution:

Question 59.
Prove that: (sin θ + cos θ + 1). (sin θ – 1 + cos θ) . sec θ . cosec θ = 2 (2014)
Solution:
L.H.S. = (sin θ + cos θ + 1) (sin θ – 1 + cos θ) . sec θ cosec θ
= [(sin θ + cos θ) + 1] [(sin θ + cos θ) – 1] . sec θ cosec θ
= [(sin θ + cos θ)2 – (1)2] sec θ cosec θ …[∵ (a + b)(a – b) = a2 – b2
= (sin2 θ + cos2θ + 2 sin θ cos θ – 1]. sec θ cosec θ
= (1 + 2 sin θ cos θ – 1). sec θ cosecθ …[∵ sin2θ + cos2θ = 1
= (2 sin θ cos θ). $$\frac{1}{\cos \theta} \cdot \frac{1}{\sin \theta}$$
= 2 = R.H.S. …(Hence proved)

Question 60.
Prove that: (2014)

Solution:

Question 61.
Prove that: (1 + cot A + tan A). (sin A – cos A) = $$\frac{\sec ^{3} A-\csc ^{3} A}{\sec ^{2} A \cdot \csc ^{2} A}$$ (2015)
Solution:

Question 62.
Prove the identity: (2015)

Solution:

Question 63.
Prove the following trigonometric identities: sin A (1 + tan A) + cos A (1 + cot A) = sec A + cosec A. (2015)
Solution:
L.H.S.
= sin A (1 + tan A) + cos A (1 + cot A)

Question 64.
Prove that: (cot A + sec B)2 – (tan B – cosec A)2 = 2(cot A . sec B + tan B. cosec A) (2014)
Solution:
L.H.S.
= (cot A + sec B)2 – (tan B – cosec A)2
= cot2 A + sec2 B + 2 cot A sec B – (tan2 B + cosec2 A – 2 tan B cosec A)
= cot2 A + sec2 B + 2 cot A sec B – tan2 B – cosec2 A + 2 tan B cosec A
= (sec2 B – tan2 B) – (cosec2 A – cot2 A) + 2(cot A sec B + tan B cosec A)
= 1 – 1 + 2(cot A sec B + tan B cosec A) … [∵ sec2B – tan2 B = 1
cosec2A – cot2 A = 1
= 2(cot A . sec B + tan B . cosec A) = R.H.S.

Question 65.
If x = r sin A cos C, y = r sin A sin C and z = r cos A, then prove that x2 + y2 + z2 = r2. (2017OD)
Solution:
x = r sin A cos C; y = r sin A sin C; z = r cos A
L.H.S. x2 + y2 + z2 = 2 sin2 A cos2C + r2 sin2 A sin2 C + r2 cos2 A
= r2 sin2 A(cos2 C + sin2 C) + r2 cos2 A
= r2 sin2 A + r2 cos2 A … [cos2θ + sin2θ = 1
= r2 (sin2 A + cos2 A) = r2 = R.H.S.

Question 66.
Prove that: (2017OD)

Solution:

Question 67.
In the adjoining figure, ABCD is a rectanlge with breadth BC = 7 cm and ∠CAB = 30°. Find the length of side AB of the rectangle and length of diagonal AC. If the ∠CAB = 60°, then what is the size of the side AB of the rectangle. [Use $$\sqrt{3}$$ = 1.73 and $$\sqrt{2}$$ = 1.41, if required) (2014OD)

Solution:

## Related content

 NCERT Exemplar for Class 6 Maths Solutions CBSE Notes for Class 8 Maths CBSE Notes for Class 7 Science CBSE Notes for Class 8 Science Lines and Angles Class 9 Extra Questions Maths Chapter 6 AMU Class 11 Entrance Exam Sample Papers Harappan Civilization Latitude and Longitude Sahara Desert Home Rule Movement

+91

Live ClassesBooksTest SeriesSelf Learning

Verify OTP Code (required)