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

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

RD Sharma Solutions for Class 10 Maths Chapter 6 – Trigonometric Identities is provided here. The branch of Mathematics which deals with the measurement of the sides and the angles of a triangle is known as trigonometry. Students who find difficulty to understand the concepts covered in this chapter can make their learning process smooth and hassle-free using RD Sharma Solutions. These solutions are well structured by our expert team at BYJUâ€™S to help students grasp the in-depth knowledge of concepts which are vital for examinations.

Trigonometric Identities is the 6th Chapter of RD Sharma Solutions Class 10. This chapter consists of two exercises. Students can find the precise answers for these exercises in RD Sharma Solutions for Class 10. The previous chapter was about trigonometric ratios and relations between them. But this chapter explains the trigonometric identities in a comprehensive manner in accordance with the student’s intelligence quotient.

## Download the PDF of RD Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities here

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

Prove the following trigonometric identities:

1. (1 â€“ cos2 A) cosec2 A = 1

Solution:

Taking the L.H.S,

(1 â€“ cos2 A) cosec2 A

= (sin2 A) cosec2 A [âˆµ sin2 A + cos2 A = 1 â‡’1 â€“ sin2 A = cos2 A]

= 12

= 1 = R.H.S

– Hence Proved

2. (1 + cot2Â A) sin2Â A = 1Â

Solution:Â

By using the identity,

cosec2 A – cot2Â A = 1 â‡’ cosec2 A = cot2Â A + 1

Taking,

L.H.S = (1 + cot2Â A) sin2Â A

= cosec2Â A sin2Â A

= (cosec A sin A)2

= ((1/sin A) Ã— sin A)2

= (1)2

= 1

= R.H.S

– Hence Proved

3. tan2 Î¸ cos2 Î¸ =Â 1 âˆ’ cos2 Î¸Â

Solution:Â

We know that,

sin2Â Î¸ + cos2Â Î¸ = 1

Taking,

L.H.S =Â tan2Â Î¸ cos2Â Î¸

= (tan Î¸ Ã— cos Î¸)2

= (sin Î¸)2

= sin2 Î¸

= 1 – cos2 Î¸

= R.H.S

– Hence Proved

4. cosec Î¸ âˆš(1 â€“ cos2 Î¸) = 1

Solution:

Using identity,

sin2Â Î¸ + cos2Â Î¸ = 1Â  â‡’ sin2Â Î¸ = 1 – cos2Â Î¸

Taking L.H.S,

L.H.S = cosec Î¸ âˆš(1 â€“ cos2 Î¸)

= cosec Î¸ âˆš( sin2Â Î¸)

= cosec Î¸ x sinÂ Î¸

= 1

= R.H.S

– Hence Proved

5. (sec2 Î¸ âˆ’ 1)(cosec2 Î¸ âˆ’ 1) = 1Â

Solution:

Using identities,

(sec2 Î¸ âˆ’ tan2 Î¸) = 1Â and (cosec2 Î¸ âˆ’ cot2 Î¸) = 1

We have,

L.H.S =Â (sec2 Î¸ – 1)(cosec2Î¸ – 1)

= tan2Î¸ Ã— cot2Î¸

= (tan Î¸ Ã— cot Î¸)2

= (tan Î¸ Ã— 1/tan Î¸)2

= 12

= 1

= R.H.S

– Hence Proved

6. tan Î¸ + 1/ tan Î¸ = sec Î¸ cosec Î¸

Solution:

We have,

L.H.S = tan Î¸ + 1/ tan Î¸

= (tan2 Î¸ + 1)/ tan Î¸

= sec2 Î¸ / tan Î¸ [âˆµ sec2 Î¸ âˆ’ tan2 Î¸ = 1]

= (1/cos2 Î¸) x 1/ (sin Î¸/cos Î¸) [âˆµ tan Î¸ = sin Î¸ / cos Î¸]

= cos Î¸/ (sin Î¸ x cos2 Î¸)

= 1/ cos Î¸ x 1/ sin Î¸

= sec Î¸ x cosec Î¸

= sec Î¸ cosec Î¸

= R.H.S

– Hence Proved

7. cos Î¸/ (1 – sin Î¸) = (1 + sin Î¸)/ cos Î¸

Solution:

We know that,

sin2 Î¸ + cos2 Î¸ = 1

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

L.H.S = R.H.S

– Hence Proved

8. cos Î¸/ (1 + sin Î¸) = (1 – sin Î¸)/ cos Î¸

Solution:

We know that,

sin2 Î¸ + cos2 Î¸ = 1

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

L.H.S = R.H.S

– Hence Proved

9. cos2 Î¸ + 1/(1 + cot2 Î¸) = 1

Solution:

cosec2 Î¸ âˆ’ cot2 Î¸ = 1 and sin2 Î¸ + cos2 Î¸ = 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:

sec2 Î¸ âˆ’ tan2 Î¸ = 1Â and sin2 Î¸ + cos2 Î¸ = 1

Taking L.H.S,

= sin2 A + cos2 A

= 1

= R.H.S

– Hence Proved

11.

Solution:

We know that,

sin2 Î¸ + cos2 Î¸ = 1

Taking the L.H.S,

= cosec Î¸ – cotÂ Î¸

= R.H.S

– Hence Proved

12. 1 â€“ cos Î¸/ sin Î¸ = sin Î¸/ 1 + cos Î¸

Solution:

We know that,

sin2 Î¸ + cos2 Î¸ = 1

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

= R.H.S

– Hence Proved

13. sin Î¸/ (1 â€“ cos Î¸) = cosec Î¸ + cot Î¸

Solution:

Taking L.H.S,

= cosec Î¸ + cot Î¸

= R.H.S

– Hence Proved

14. (1 â€“ sin Î¸) / (1 + sin Î¸) = (sec Î¸ â€“ tan Î¸)2

Solution:

Taking the L.H.S,

= (sec Î¸ – tan Î¸)2

= R.H.S

– Hence Proved

15.

Solution:

Taking L.H.S,

= cot Î¸

= R.H.S

– Hence Proved

16. tan2 Î¸ âˆ’ sin2 Î¸Â =Â tan2 Î¸ sin2 Î¸Â

Solution:

Taking L.H.S,

L.H.S = tan2 Î¸ âˆ’ sin2 Î¸Â

= tan2 Î¸ sin2 Î¸

= R.H.S

– Hence Proved

17. (cosecÂ Î¸Â + sinÂ Î¸)(cosecÂ Î¸Â – sinÂ Î¸) =Â cot2Î¸ + cos2Î¸Â

Solution:

Taking L.H.S = (cosecÂ Î¸Â + sinÂ Î¸)(cosecÂ Î¸Â – sinÂ Î¸)

On multiplying we get,

= cosec2Â Î¸Â â€“ sin2Â Î¸

=Â (1 + cot2Â Î¸)Â – (1 – cos2Â Î¸) [Using cosec2 Î¸ âˆ’ cot2 Î¸ = 1Â and sin2 Î¸ + cos2 Î¸ = 1]

=Â 1 + cot2Â Î¸Â – 1 + cos2Â Î¸

= cot2Â Î¸Â + cos2Â Î¸

= R.H.S

– Hence Proved

18. (sec Î¸Â + cosÂ Î¸) (secÂ Î¸Â – cosÂ Î¸) =Â tan2 Î¸ + sin2 Î¸Â

Solution:

Taking L.H.S = (secÂ Î¸Â + cosÂ Î¸)(secÂ Î¸Â – cosÂ Î¸)

On multiplying we get,

= sec2Â Î¸Â â€“ sin2Â Î¸

= (1 + tan2Â Î¸)Â – (1 – sin2Â Î¸) [Using sec2 Î¸ âˆ’ tan2 Î¸ = 1Â and sin2 Î¸ + cos2 Î¸ = 1]

= 1 + tan2Â Î¸Â – 1 + sin2Â Î¸

=Â tan 2 Î¸Â + sin 2Â Î¸

= R.H.S

– Hence Proved

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

Solution:

Taking L.H.S = sec A(1 â€“ 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 â€“ sin A)(1/cos A + sin A/cos A)

= 1 â€“ sin2 A / cos2 A [After taking L.C.M]

= cos2 A / cos2 A [âˆµ 1 â€“ sin2 A = cos2 A]

= 1

= R.H.S

– Hence Proved

20. (cosec A â€“ sin A)(sec A â€“ cos A)(tan A + cot A) = 1Â

Solution:

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

= (cos2 A/ sin A) (sin2 A/ cos A) (1/ sin A cos A) [âˆµ sin2 Î¸ + cos2 Î¸ = 1]

= (sin AÂ  cos A)Â  (1/ cos A sin A)

= 1

= R.H.S

– Hence Proved

21. (1 +Â tan2 Î¸)(1 – sinÂ Î¸)(1 + sinÂ Î¸) = 1

Solution:

Taking L.H.S = (1 +Â tan2Î¸)(1 – sin Î¸)(1 + sin Î¸)

And, we know sin2Â Î¸Â + cos2Â Î¸Â = 1 and sec2Â Î¸Â – tan2Â Î¸Â = 1

So,

L.H.S = (1 +Â tan2 Î¸)(1 – sinÂ Î¸)(1 + sinÂ Î¸)

= (1 +Â tan2 Î¸){(1 – sinÂ Î¸)(1 + sinÂ Î¸)}

= (1 +Â tan2 Î¸)(1 –Â sin2 Î¸)

= sec2 Î¸ (cos2 Î¸)

= (1/ cos2 Î¸) x cos2 Î¸

= 1

= R.H.S

– Hence Proved

22. sin2 A cot2 A + cos2 A tan2 AÂ = 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 [âˆµ sin2 Î¸ + cos2 Î¸ = 1]

= R.H.S

– Hence Proved

23.

Solution:

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

L.H.S = cot Î¸ â€“ tan Î¸

= R.H.S

– Hence Proved

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

L.H.S = tan Î¸ â€“ cot Î¸

= R.H.S

– Hence Proved

24. (cos2 Î¸/ sin Î¸) â€“ cosec Î¸ + sin Î¸ = 0

Solution:

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

= â€“ sin Î¸ + sin Î¸

= 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 Î¸ + cos2 Î¸ = 1, we have

= 2/ cos Î¸

= 2 sec Î¸

= R.H.S

– Hence proved

27.

Solution:

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

= R.H.S

• Hence proved

28.

Solution:

Taking L.H.S,

Using sec2 Î¸ âˆ’ tan2 Î¸ = 1Â and cosec2 Î¸ âˆ’ cot2 Î¸ = 1

= R.H.S

– Hence proved

29.

Solution:

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

= R.H.S

– Hence proved

30.

Solution:

Taking LHS, we have

= 1 + tan Î¸ + cot Î¸

= R.H.S

– Hence proved

31. sec6 Î¸ = tan6 Î¸ + 3 tan2Â Î¸ sec2 Î¸ + 1

Solution:Â

From trig. Identities we have,

sec2 Î¸ âˆ’ tan2 Î¸ = 1

On cubingÂ both sides,

(sec2Î¸ âˆ’ tan2Î¸)3 = 1

sec6 Î¸ âˆ’ tan6 Î¸ âˆ’ 3sec2 Î¸ tan2 Î¸(sec2 Î¸ âˆ’ tan2 Î¸) = 1

[Since,Â (aÂ â€“ b)3Â = a3Â – b3 â€“ 3ab(a – b)]

sec6 Î¸ âˆ’ tan6 Î¸ âˆ’ 3sec2 Î¸ tan2 Î¸ = 1

â‡’ sec6 Î¸ = tan6 Î¸ + 3sec2 Î¸ tan2 Î¸ + 1

L.H.S = R.H.S

– Hence proved

32. cosec6 Î¸ = cot6 Î¸ + 3cot2 Î¸ cosec2 Î¸ + 1

Solution:

From trig. Identities we have,

cosec2 Î¸ âˆ’ cot2 Î¸ = 1

On cubingÂ both sides,

(cosec2 Î¸ âˆ’ cot2 Î¸)3Â = 1

cosec6 Î¸ âˆ’ cot6 Î¸ âˆ’ 3cosec2 Î¸ cot2 Î¸ (cosec2 Î¸ âˆ’ cot2 Î¸) = 1

[Since,Â (aÂ â€“ b)3Â = a3Â – b3 â€“ 3ab(a – b)]

cosec6 Î¸ âˆ’ cot6 Î¸ âˆ’ 3cosec2 Î¸ cot2 Î¸ = 1

â‡’ cosec6 Î¸ = cot6 Î¸ + 3 cosec2 Î¸ cot2 Î¸ + 1

L.H.S = R.H.S

– Hence proved

33.

Solution:

Taking L.H.S and using sec2 Î¸ âˆ’ tan2 Î¸ = 1Â â‡’ 1 + tan2 Î¸ = sec2 Î¸

= R.H.S

– Hence proved

34.

Solution:

Taking L.H.S and using the identity sin2A + cos2AÂ = 1, we get

sin2A = 1 âˆ’ cos2A

â‡’ sin2A = (1 â€“ cos A)(1 + cos A)

– Hence proved

35.

Solution:

We have,

Rationalizing the denominator and numerator with (sec A + tan A) and using sec2 Î¸ âˆ’ tan2 Î¸ = 1Â we get,

= R.H.S

– Hence proved

36.

Solution:

We have,

On multiplying numerator and denominator by (1 â€“ cos A), we get

= R.H.S

– Hence proved

37. (i)

Solution:

Taking L.H.S and rationalizing the numerator and denominator with âˆš(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 Î¸

= 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 Î¸

= R.H.S

– Hence proved

(iv)

Solution:

Taking L.H.S, we have

= R.H.S

– Hence proved

39.

Solution:

Taking LHS = (sec A â€“ tan A)2 , we have

= R.H.S

– Hence proved

40.

Solution:

Taking L.H.S and rationalizing the numerator and denominator with (1 â€“ 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. IfÂ cos Î¸ = 4/5, find all other trigonometric ratios of angleÂ Î¸.Â

Solution:

We have,

cos Î¸ = 4/5

And we know that,

sin Î¸ = âˆš(1 – cos2 Î¸)

â‡’ sin Î¸ = âˆš(1 â€“ (4/5)2)

= âˆš(1 â€“ (16/25))

= âˆš[(25 â€“ 16)/25]

= âˆš(9/25)

= 3/5

âˆ´ sin Î¸ = 3/5

Since, cosec Î¸ = 1/ sin Î¸

= 1/ (3/5)

â‡’ cosec Î¸ = 5/3

And, sec Î¸ = 1/ cos Î¸

= 1/ (4/5)

â‡’ cosec Î¸ = 5/4

Now,

tan Î¸ = sin Î¸/ cos Î¸

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

â‡’ tan Î¸ = 3/4

And, cot Î¸ = 1/ tan Î¸

= 1/ (3/4)

â‡’ cot Î¸ = 4/3

2. IfÂ sin Î¸ = 1/âˆš2, find all other trigonometric ratios of angleÂ Î¸.

Solution:

We have,

sin Î¸ = 1/âˆš2

And we know that,

cos Î¸ = âˆš(1 – sin2 Î¸)

â‡’ cos Î¸ = âˆš(1 â€“ (1/âˆš2)2)

= âˆš(1 â€“ (1/2))

= âˆš[(2 â€“ 1)/2]

= âˆš(1/2)

= 1/âˆš2

âˆ´ cos Î¸ = 1/âˆš2

Since, cosec Î¸ = 1/ sin Î¸

= 1/ (1/âˆš2)

â‡’ cosec Î¸ = âˆš2

And, sec Î¸ = 1/ cos Î¸

= 1/ (1/âˆš2)

â‡’ sec Î¸ = âˆš2

Now,

tan Î¸ = sin Î¸/ cos Î¸

= (1/âˆš2)/ (1/âˆš2)

â‡’ tan Î¸ = 1

And, cot Î¸ = 1/ tan Î¸

= 1/ (1)

â‡’ cot Î¸ = 1

3.

Solution:

Given,

tan Î¸ = 1/âˆš2

By using sec2 Î¸ âˆ’ tan2 Î¸ = 1,

4.

Solution:

Given,

tan Î¸ = 3/4

By using sec2 Î¸ âˆ’ tan2 Î¸ = 1,

sec Î¸ = 5/4

Since, sec Î¸ = 1/ cos Î¸

â‡’ cos Î¸ = 1/ sec Î¸

= 1/ (5/4)

= 4/5

So,

$$\begin{array}{l}\frac{1-cos\theta }{1+cos\theta }=\frac{1-\frac{4}{5}}{1+\frac{4}{5}}=\frac{\frac{1}{5}}{\frac{9}{5}}=\frac{1}{9}\end{array}$$

5.Â

Solution:

Given, tan Î¸ = 12/5

Since, cot Î¸ = 1/ tan Î¸ = 1/ (12/5) = 5/12

Now, by using cosec2 Î¸ âˆ’ cot2 Î¸ = 1

cosec Î¸ = âˆš(1 + cot2 Î¸)

= âˆš(1 + (5/12)2 )

= âˆš(1 + 25/144)

= âˆš(169/ 25)

â‡’ cosec Î¸ = 13/5

Now, we know that

sin Î¸ = 1/ cosec Î¸

= 1/ (13/5)

â‡’ sin Î¸ = 5/13

Putting value of sin Î¸ in the expression we have,

= 25/ 1

= 25

6.

Solution:

Given,

cot Î¸ = 1/âˆš3

Using cosec2 Î¸ âˆ’ cot2 Î¸ = 1, we can find cosec Î¸

cosec Î¸ = âˆš(1 + cot2 Î¸)

= âˆš(1 + (1/âˆš3)2)

= âˆš(1 + (1/3)) = âˆš((3 + 1)/3)

= âˆš(4/3)

â‡’ cosec Î¸ = 2/âˆš3

So, sin Î¸ = 1/ cosec Î¸ = 1/ (2/âˆš3)

â‡’ sin Î¸ = âˆš3/2

And, we know that

cos Î¸ = âˆš(1 – sin2 Î¸)

= âˆš(1 â€“ (âˆš3/2)2)

= âˆš(1 â€“ (3/4))

= âˆš((4 â€“ 3)/4)

= âˆš(1/4)

â‡’ cos Î¸ = 1/2

Now, using cos Î¸ and sin Î¸ in the expression, we have

= 3/5

7.

Solution:

Given,

cosec A = âˆš2

Using cosec2 A âˆ’ cot2 A = 1, we find cot A

= 4/2

= 2

## Frequently Asked Questions on RD Sharma Solutions for Class 10 Maths Chapter 6

### Why should we follow RD Sharma Solutions for Class 10 Maths Chapter 6 during exam preparation?

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### Is RD Sharma Solutions for Class 10 Maths Chapter 6 important from an exam point of view?

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