RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities

RD Sharma Solutions for Class 10 Maths Chapter 6 – Trigonometric Identities are an essential resource for students aiming to excel in mathematics. This chapter is part of the RD Sharma Class 10 textbook, which is widely recognized for its comprehensive approach to learning. Students can also access the RD Sharma Class 10 new edition PDF for updated content and exercises. 

Trigonometry, a branch of mathematics that deals with the measurement of triangle sides and angles, is thoroughly explained in this chapter. For those who find these concepts challenging, RD Sharma Solutions for Class 10 Maths provide a structured and simplified learning experience. These RD Sharma solutions, prepared by expert educators at Infinity Learn, help students grasp critical concepts vital for exams.

Chapter 6 – Trigonometric Identities includes two exercises, and students can find precise answers to these problems in the solutions provided. Additionally, students seeking extra practice can explore Trigonometry questions for Class 10 PDF download for targeted preparation. 

While the previous chapter focused on trigonometric ratios and their relationships, this chapter offers an in-depth explanation of trigonometric identities tailored to varying levels of understanding. 

For convenient access, students can opt for the RD Sharma Class 10 textbook PDF free download, ensuring they have all necessary materials at their fingertips.

RD Sharma Solutions Class 10 Maths Chapter 6 – Free PDF Download

RD Sharma Class 10 Chapter 6 PDF includes detailed solutions, examples, and extra questions to help you master trigonometric identities and other topics. Click here to download the RD Sharma Class 10 Chapter 6 PDF.

Access the RD Sharma Solutions for Class 10 Maths Chapter 6 – Trigonometric Identities

Prove the following trigonometric identities:

Q. (1 + cot² A) sin² A = 1

Solution:

By using the identity,

cosec² A – cot² A = 1 

⇒ cosec² A 

= cot2 A + 1

Taking,

L.H.S. = (1 + cot2 A) sin2 A

= cosec2 A sin2 A

= (cosec A sin A)²

= ((1/sin A) × sin A)²

= (1)²

= 1

= R.H.S.

– Hence, proved.

Q. (sec² θ − 1)(cosec² θ − 1) = 1

Solution:

Using identities,

(sec² θ − tan² θ) = 1 and (cosec² θ − cot² θ) = 1

We have,

L.H.S. = (sec² θ – 1)(cosec²θ – 1)

= tan²θ × cot²θ

= (tan θ × cot θ)²

= (tan θ × 1/tan θ)²

= 1²

= 1

= R.H.S.

Hence, proved.

Q. (cosec θ + sin θ)(cosec θ – sin θ) = cot²θ + cos²θ

Solution:

Taking L.H.S. = (cosec θ + sin θ)(cosec θ – sin θ)

On multiplying, we get

= cosec² θ – sin²θ

= (1 + cot²θ) – (1 – cos²θ) [Using cosec²θ − cot²θ = 1 and sin²θ + cos²θ = 1]

= 1 + cot²θ – 1 + cos²θ

= cot²θ + cos²θ

= R.H.S.

Hence, proved.

Q. sec A(1- sin A) (sec A + tan A) = 1

Solution:

Taking L.H.S. = sec A(1 – sin A)(sec A + tan A)

Substituting sec A = 1/cos A and tan A =sin A/cos A in the above, we have,

L.H.S = 1/cos A (1 – sin A)(1/cos A + sin A/cos A)

= 1 – sin²A / cos²A [After taking L.C.M]

= cos²A / cos²A   [1 – sin² A = cos²A]

= 1

= R.H.S.

Q. (1 + tan²θ)(1 – sin θ)(1 + sin θ) = 1

Solution:

Taking L.H.S. = (1 + tan2θ)(1 – sin θ)(1 + sin θ)

And, we know sin²θ + cos²θ = 1 and sec²θ – tan²θ = 1

So,

L.H.S = (1 + tan²θ)(1 – sin θ)(1 + sin θ)

= (1 + tan²θ){(1 – sin θ)(1 + sin θ)}

= (1 + tan²θ)(1 – sin²θ)

= sec²θ (cos²θ)

= (1/ cos²θ) x cos²θ

= 1

= R.H.S.

Q. If sin θ = 1/√2, find all other trigonometric ratios of angle θ.

Solution:

We have,

sin θ = 1/√2

And we know that,

cos θ = √(1 – sin2 θ)

⇒ cos θ = √(1 – (1/√2)²)

= √(1 – (1/2))

= √[(2 – 1)/2]

= √(1/2)

= 1/√2

∴ cos θ = 1/√2

Since, cosec θ = 1/ sin θ

= 1/ (1/√2)

⇒ cosec θ = √2

And, sec θ = 1/ cos θ

= 1/ (1/√2)

= sec θ = √2

Now,

tan θ = sin θ/ cos θ

= (1/√2)/ (1/√2)

⇒ tan θ = 1

And, cot θ = 1/ tan θ

= 1/ (1)

⇒ cot θ = 1