# RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities

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.

### 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 + cosA = 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,

cosecA – cot2 A = 1 ⇒ cosecA = 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. tanθ cosθ = 1 − cosθ

Solution:

We know that,

sinθ + cosθ = 1

Taking,

L.H.S = tanθ cosθ

= (tan θ × cos θ)2

= (sin θ)2

= sinθ

= 1 – cosθ

= R.H.S

– Hence Proved

4. cosec θ √(1 – cos2 θ) = 1

Solution:

Using identity,

sinθ + cosθ = 1  ⇒ sinθ = 1 – cosθ

Taking L.H.S,

L.H.S = cosec θ √(1 – cos2 θ)

= cosec θ √( sinθ)

= cosec θ x sin θ

= 1

= R.H.S

– Hence Proved

5. (secθ − 1)(cosecθ − 1) = 1

Solution:

Using identities,

(secθ − tanθ) = 1 and (cosecθ − cotθ) = 1

We have,

L.H.S = (secθ – 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 [∵ secθ − tanθ = 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,

sinθ + cosθ = 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,

sinθ + cosθ = 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. cosθ + 1/(1 + cotθ) = 1

Solution:

cosecθ − cotθ = 1 and sinθ + cosθ = 1

Taking L.H.S, = cos2 A + sin2 A

= 1

= R.H.S

– Hence Proved

10. sinA + 1/(1 + tan A) = 1

Solution:

secθ − tanθ = 1 and sinθ + cosθ = 1

Taking L.H.S, = sin2 A + cos2 A

= 1

= R.H.S

– Hence Proved

11. Solution:

We know that, sinθ + cosθ = 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,

sinθ + cosθ = 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. tanθ − sinθ = tanθ sinθ

Solution:

Taking L.H.S,

L.H.S = tanθ − sinθ = tanθ sinθ

= 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 cosecθ − cotθ = 1 and sinθ + cosθ = 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) = tanθ + sinθ

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 secθ − tanθ = 1 and sinθ + cosθ = 1]

= 1 + tan2 θ – 1 + sin2 θ

= tan 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) [∵ sinθ + cosθ = 1]

= (sin A  cos A)  (1/ cos A sin A)

= 1

= R.H.S

– Hence Proved

21. (1 + tanθ)(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 + tanθ){(1 – sin θ)(1 + sin θ)}

= (1 + tanθ)(1 – sinθ)

= secθ (cos2 θ)

= (1/ cos2 θ) x cos2 θ

= 1

= R.H.S

– Hence Proved

22. sinA cotA + cosA tanA = 1

Solution:

We know that,

cotA = cosA/ sin2 A and tan2 A = sin2 A/cos2 A

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

L.H.S = sinA cotA + cosA tanA

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

= cosA + sin2 A

= 1 [∵ sinθ + cosθ = 1]

= R.H.S

– Hence Proved 23.

Solution:

(i) Taking the L.H.S and using sinθ + cosθ = 1, we have

L.H.S = cot θ – tan θ = R.H.S

– Hence Proved

(ii) Taking the L.H.S and using sinθ + cosθ = 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 sinθ + cosθ = 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 sinθ + cosθ = 1, we have = 2/ cos θ

= 2 sec θ

= R.H.S

• Hence proved 27.

Solution:

Taking the LHS and using sinθ + cosθ = 1, we have = R.H.S

• Hence proved

28. Solution:

Taking L.H.S, Using secθ − tanθ = 1 and cosecθ − cotθ = 1 = R.H.S

• Hence proved 29.

Solution:

Taking L.H.S and using sinθ + cosθ = 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. secθ = tanθ + 3 tanθ secθ + 1

Solution:

From trig. Identities we have,

secθ − tanθ = 1

On cubing both sides,

(sec2θ − tan2θ)= 1

secθ − tanθ − 3secθ tanθ(secθ − tanθ) = 1

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

secθ − tanθ − 3secθ tanθ = 1

⇒ secθ = tanθ + 3secθ tanθ + 1

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

• Hence proved

32. cosecθ = cotθ + 3cotθ cosecθ + 1

Solution:

From trig. Identities we have,

cosecθ − cotθ = 1

On cubing both sides,

(cosecθ − cotθ)3 = 1

cosecθ − cotθ − 3cosecθ cotθ (cosecθ − cotθ) = 1

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

cosecθ − cotθ − 3cosecθ cotθ = 1

⇒ cosecθ = cotθ + 3 cosecθ cotθ + 1

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

• Hence proved 33.

Solution:

Taking L.H.S and using secθ − tanθ = 1 ⇒ 1 + tanθ = secθ = 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 secθ − tanθ = 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 secA

= 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 – cosθ)

⇒ 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 – sinθ)

⇒ 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 secθ − tanθ = 1,  4.

Solution:

Given,

tan θ = 3/4

By using secθ − tanθ = 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 cosecθ − cotθ = 1

cosec θ = √(1 + cotθ)

= √(1 + (5/12))

= √(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 cosecθ − cotθ = 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 cosecA − cotA = 1, we find cot A = 4/2

= 2

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