The branch of mathematics which deals with the measurement of the sides and the angles of a triangle is trigonometry. We know that by now this topic would already seem difficult and complicated as its completely new to you. So, in order to make your learning process smooth and hassle-free the RD Sharma Solutions prepared by Goyanka Maths Study will help students get the correct understanding of various chapters in the book.

Trigonometric Identities is the 6th chapter of RD Sharma Class 10 which has two exercises and its solved answers with detailed explanations are given here RD Sharma Solutions for Class 10. The previous chapter was about trigonometric ratios and relations between them. But this chapter will be about proving some trigonometric identities and use them to prove other useful trigonometric identities.

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

### RD Sharma Solutions for Class 10 Maths Chapter 6 Exercise 6.1 Page No: 6.43

**Prove the following trigonometric identities:**

**1. (1 – cos2 A) cosec2 A = 1**

**Solution:**

Taking the L.H.S,

(1 – cos2 A) cosec2 A

= (sin2 A) cosec2 A [∵ sin2 A + cos2 A = 1 ⇒1 – sin2 A = cos2 A]

= 12

= 1 = R.H.S

– Hence Proved

**2. (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

**3. tan2 θ cos2 θ = 1 − cos2 θ **

**Solution: **

We know that,

sin2 θ + cos2 θ = 1

Taking,

L.H.S = tan2 θ cos2 θ

= (tan θ × cos θ)2

= (sin θ)2

= sin2 θ

= 1 – cos2 θ

= R.H.S

– Hence Proved

**4. cosec θ √(1 – cos2 θ) = 1**

**Solution:**

Using identity,

sin2 θ + cos2 θ = 1 ⇒ sin2 θ = 1 – cos2 θ

Taking L.H.S,

L.H.S = cosec θ √(1 – cos2 θ)

= cosec θ √( sin2 θ)

= cosec θ x sin θ

= 1

= R.H.S

– Hence Proved

**5. (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

**6. tan θ + 1/ tan θ = sec θ cosec θ**

**Solution:**

We have,

L.H.S = tan θ + 1/ tan θ

= (tan2 θ + 1)/ tan θ

= sec2 θ / tan θ [∵ sec2 θ − tan2 θ = 1]

= (1/cos2 θ) x 1/ (sin θ/cos θ) [∵ tan θ = sin θ / cos θ]

= cos θ/ (sin θ x cos2 θ)

= 1/ cos θ x 1/ sin θ

= sec θ x cosec θ

= sec θ cosec θ

= R.H.S

– Hence Proved

**7. cos θ/ (1 – sin θ) = (1 + sin θ)/ cos θ**

**Solution:**

We know that,

sin2 θ + cos2 θ = 1

So, by multiplying both the numerator and the denominator by (1+ sin θ), we get

L.H.S =

= R.H.S

– Hence Proved

8. **cos θ/ (1 + sin θ) = (1 – sin θ)/ cos θ**

**Solution:**

We know that,

sin2 θ + cos2 θ = 1

So, by multiplying both the numerator and the denominator by (1- sin θ), we get

L.H.S =

= R.H.S

– Hence Proved

**9. cos2 θ + 1/(1 + cot2 θ) = 1**

**Solution:**

We already know that,

cosec2 θ − cot2 θ = 1 and sin2 θ + cos2 θ = 1

Taking L.H.S,

= cos2 A + sin2 A

= 1

= R.H.S

– Hence Proved

**10. sin2 A + 1/(1 + tan 2 A) = 1**

**Solution:**

We already know that,

sec2 θ − tan2 θ = 1 and sin2 θ + cos2 θ = 1

Taking L.H.S,

= sin2 A + cos2 A

= 1

= R.H.S

– Hence Proved

**11.**

**Solution:**

We know that, sin2 θ + cos2 θ = 1

Taking the L.H.S,

= cosec θ – cot θ

= R.H.S

– Hence Proved

**12. 1 – cos θ/ sin θ = sin θ/ 1 + cos θ**

**Solution:**

We know that,

sin2 θ + cos2 θ = 1

So, by multiplying both the numerator and the denominator by (1+ cos θ), we get

= R.H.S

– Hence Proved

**13.** **sin θ/ (1 – cos θ) = cosec θ + cot θ**

**Solution:**

Taking L.H.S,

= cosec θ + cot θ

= R.H.S

– Hence Proved

**14. (1 – sin θ) / (1 + sin θ) = (sec θ – tan θ)2**

**Solution:**

Taking the L.H.S,

= (sec θ – tan θ)2

= R.H.S

– Hence Proved

**15. **

**Solution:**

Taking L.H.S,

= cot θ

= R.H.S

– Hence Proved

**16. tan2 θ − sin2 θ = tan2 θ sin2 θ **

**Solution:**

Taking L.H.S,

L.H.S = tan2 θ − sin2 θ** **

= tan2 θ sin2 θ

= R.H.S

– Hence Proved

**17. (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 – sin2 θ

= (1 + cot2 θ) – (1 – cos2 θ) [Using cosec2 θ − cot2 θ = 1 and sin2 θ + cos2 θ = 1]

= 1 + cot2 θ – 1 + cos2 θ

= cot2 θ + cos2 θ

= R.H.S

– Hence Proved

**18. (sec θ + cos θ) (sec θ – cos θ) = tan2 θ + sin2 θ **

**Solution:**

Taking L.H.S = (sec θ + cos θ)(sec θ – cos θ)

On multiplying we get,

= sec2 θ – sin2 θ

= (1 + tan2 θ) – (1 – sin2 θ) [Using sec2 θ − tan2 θ = 1 and sin2 θ + cos2 θ = 1]

= 1 + tan2 θ – 1 + sin2 θ

= tan 2 θ + sin 2 θ

= R.H.S

– Hence Proved

**19. 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 – sin2 A / cos2 A [After taking L.C.M]

= cos2 A / cos2 A [∵ 1 – sin2 A = cos2 A]

= 1

= R.H.S

– Hence Proved

**20. (cosec A – sin A)(sec A – cos A)(tan A + cot A) = 1 **

**Solution:**

Taking L.H.S = (cosec A – sin A)(sec A – cos A)(tan A + cot A)

= (cos2 A/ sin A) (sin2 A/ cos A) (1/ sin A cos A) [∵ sin2 θ + cos2 θ = 1]

= (sin A cos A) (1/ cos A sin A)

= 1

= R.H.S

– Hence Proved

**21. (1 + tan2 θ)(1 – sin θ)(1 + sin θ) = 1**

**Solution:**

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

And, we know sin2 θ + cos2 θ = 1 and sec2 θ – tan2 θ = 1

So,

L.H.S = (1 + tan2 θ)(1 – sin θ)(1 + sin θ)

= (1 + tan2 θ){(1 – sin θ)(1 + sin θ)}

= (1 + tan2 θ)(1 – sin2 θ)

= sec2 θ (cos2 θ)

= (1/ cos2 θ) x cos2 θ

= 1

= R.H.S

– Hence Proved

**22. sin2 A cot2 A + cos2 A tan2 A = 1**

**Solution:**

We know that,

cot2 A = cos2 A/ sin2 A and tan2 A = sin2 A/cos2 A

Substituting the above in L.H.S, we get

L.H.S = sin2 A cot2 A + cos2 A tan2 A

= {sin2 A (cos2 A/ sin2 A)} + {cos2 A (sin2 A/cos2 A)}

= cos2 A + sin2 A

= 1 [∵ sin2 θ + cos2 θ = 1]

= R.H.S

– Hence Proved

**23.**

**Solution:**

(i) Taking the L.H.S and using sin2 θ + cos2 θ = 1, we have

L.H.S = cot θ – tan θ

= R.H.S

– Hence Proved

(ii) Taking the L.H.S and using sin2 θ + cos2 θ = 1, we have

L.H.S = tan θ – cot θ

= R.H.S

– Hence Proved

**24. (cos2 θ/ sin θ) – cosec θ + sin θ = 0**

**Solution:**

Taking L.H.S and using sin2 θ + cos2 θ = 1, we have

= – sin θ + sin θ

= 0

= R.H.S

- Hence proved

**25.**

**Solution:**

Taking L.H.S,

= 2 sec2 A

= R.H.S

- Hence proved

**26. **

**Solution:**

Taking the LHS and using sin2 θ + cos2 θ = 1, we have

= 2/ cos θ

= 2 sec θ

= R.H.S

- Hence proved

**27.**

**Solution:**

Taking the LHS and using sin2 θ + cos2 θ = 1, we have

= R.H.S

- Hence proved

**28.**

**Solution:**

Taking L.H.S,

Using sec2 θ − tan2 θ = 1 and cosec2 θ − cot2 θ = 1

= R.H.S

- Hence proved

**29.**

**Solution:**

Taking L.H.S and using sin2 θ + cos2 θ = 1, we have

= R.H.S

- Hence proved

**30.**

**Solution:**

Taking LHS, we have

= 1 + tan θ + cot θ

= R.H.S

- Hence proved

**31. sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1**

**Solution: **

From trig. Identities we have,

sec2 θ − tan2 θ = 1

On cubing both sides,

(sec2θ − tan2θ)3 = 1

sec6 θ − tan6 θ − 3sec2 θ tan2 θ(sec2 θ − tan2 θ) = 1

[Since, (a – b)3 = a3 – b3 – 3ab(a – b)]

sec6 θ − tan6 θ − 3sec2 θ tan2 θ = 1

⇒ sec6 θ = tan6 θ + 3sec2 θ tan2 θ + 1

Hence, L.H.S = R.H.S

- Hence proved

**32. cosec6 θ = cot6 θ + 3cot2 θ cosec2 θ + 1**

**Solution:**

From trig. Identities we have,

cosec2 θ − cot2 θ = 1

On cubing both sides,

(cosec2 θ − cot2 θ)3 = 1

cosec6 θ − cot6 θ − 3cosec2 θ cot2 θ (cosec2 θ − cot2 θ) = 1

[Since, (a – b)3 = a3 – b3 – 3ab(a – b)]

cosec6 θ − cot6 θ − 3cosec2 θ cot2 θ = 1

⇒ cosec6 θ = cot6 θ + 3 cosec2 θ cot2 θ + 1

Hence, L.H.S = R.H.S

- Hence proved

**33.**

**Solution:**

Taking L.H.S and using sec2 θ − tan2 θ = 1 ⇒ 1 + tan2 θ = sec2 θ

= R.H.S

- Hence proved

**34.**

**Solution:**

Taking L.H.S and using the identity sin2A + cos2A = 1, we get

sin2A = 1 − cos2A

⇒ sin2A = (1 – cos A)(1 + cos A)

- Hence proved

**35.**

**Solution:**

We have,

Rationalizing the denominator and numerator with (sec A + tan A) and using sec2 θ − tan2 θ = 1 we get,

= R.H.S

- Hence proved

**36.**

**Solution:**

We have,

On multiplying numerator and denominator by (1 – cos A), we get

= R.H.S

- Hence proved

**37. (i)**

**Solution:**

Taking L.H.S and rationalizing the numerator and denominator with √(1 + sin A), we get

= R.H.S

- Hence proved

**(ii)**

**Solution:**

Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get

= 2 cosec A

= R.H.S

- Hence proved

**38. Prove that:**

**(i)**

**Solution:**

Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get

= 2 cosec θ

= R.H.S

- Hence proved

**(ii)**

**Solution:**

Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get

= R.H.S

- Hence proved

**(iii)**

**Solution:**

Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get

= 2 cosec θ

= R.H.S

- Hence proved

**(iv)**

**Solution:**

Taking L.H.S, we have

= R.H.S

- Hence proved

**39.**

**Solution:**

Taking LHS = (sec A – tan A)2 , we have

= R.H.S

- Hence proved

**40.**

**Solution:**

Taking L.H.S and rationalizing the numerator and denominator with (1 – cos A), we get

= (cosec A – cot A)2

= (cot A – cosec A)2

= R.H.S

- Hence proved

**41.**

**Solution:**

Considering L.H.S and taking L.C.M and on simplifying we have,

= 2 cosec A cot A = RHS

- Hence proved

**42.**

**Solution:**

Taking LHS, we have

= cos A + sin A

= RHS

- Hence proved

**43.**

**Solution:**

Considering L.H.S and taking L.C.M and on simplifying we have,

= 2 sec2 A

= RHS

- Hence proved

### RD Sharma Solutions for Class 10 Maths Chapter 6 Exercise 6.2 Page No: 6.54

**1. If cos θ = 4/5, find all other trigonometric ratios of angle θ. **

**Solution:**

We have,

cos θ = 4/5

And we know that,

sin θ = √(1 – cos2 θ)

⇒ sin θ = √(1 – (4/5)2)

= √(1 – (16/25))

= √[(25 – 16)/25]

= √(9/25)

= 3/5

∴ sin θ = 3/5

Since, cosec θ = 1/ sin θ

= 1/ (3/5)

⇒ cosec θ = 5/3

And, sec θ = 1/ cos θ

= 1/ (4/5)

⇒ cosec θ = 5/4

Now,

tan θ = sin θ/ cos θ

= (3/5)/ (4/5)

⇒ tan θ = 3/4

And, cot θ = 1/ tan θ

= 1/ (3/4)

⇒ cot θ = 4/3

**2. 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

**3.**

**Solution:**

Given,

tan θ = 1/√2

By using sec2 θ − tan2 θ = 1,

**4.**

**Solution:**

Given,

tan θ = 3/4

By using sec2 θ − tan2 θ = 1,

**sec **θ = 5/4

Since, sec θ = 1/ cos θ

⇒ cos θ = 1/ sec θ

= 1/ (5/4)

= 4/5

So,

**5.**

**Solution:**

Given, tan θ = 12/5

Since, cot θ = 1/ tan θ = 1/ (12/5) = 5/12

Now, by using cosec2 θ − cot2 θ = 1

cosec θ = √(1 + cot2 θ)

= √(1 + (5/12)2 )

= √(1 + 25/144)

= √(169/ 25)

⇒ cosec θ = 13/5

Now, we know that

sin θ = 1/ cosec θ

= 1/ (13/5)

⇒ sin θ = 5/13

Putting value of sin θ in the expression we have,

= 25/ 1

= 25

**6.**

**Solution:**

**Given,**

cot θ = 1/√3

Using cosec2 θ − cot2 θ = 1, we can find cosec θ

cosec θ = √(1 + cot2 θ)

= √(1 + (1/√3)2)

= √(1 + (1/3)) = √((3 + 1)/3)

= √(4/3)

⇒ cosec θ = 2/√3

So, sin θ = 1/ cosec θ = 1/ (2/√3)

⇒ sin θ = √3/2

And, we know that

cos θ = √(1 – sin2 θ)

= √(1 – (√3/2)2)

= √(1 – (3/4))

**= **√((4 – 3)/4)

= √(1/4)

⇒ cos θ = 1/2

Now, using cos θ and sin θ in the expression, we have

= 3/5

**7.**

**Solution:**

Given,

cosec A = √2

Using cosec2 A − cot2 A = 1, we find cot A

= 4/2

= 2