## 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

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

**Prove the following trigonometric identities: **

**1. (1 â€“ cos ^{2} A) cosec^{2} A = 1 **

**Solution: **

Taking the L.H.S,

(1 â€“ cos^{2} A) cosec^{2} A

= (sin^{2} A) cosec^{2} A [âˆµ sin^{2} A + cos^{2 }A = 1 â‡’1 â€“ sin^{2} A = cos^{2} A]

= 1^{2}

*= 1 = R.H.S*

– Hence Proved

**2. (1 + cot ^{2}Â A) sin^{2}Â A = 1Â **

**Solution:Â **

By using the identity,

**cosec ^{2 }A – cot^{2}Â A = 1 â‡’ cosec^{2 }A = cot^{2}Â A + 1**

Taking,

L.H.S = (1 + cot^{2}Â A) sin^{2}Â A

= cosec^{2}Â A sin^{2}Â A

= (cosec A sin A)^{2}

= ((1/sin A) Ã— sin A)^{2}

= (1)^{2}

*= 1*

= R.H.S

– Hence Proved

**3. tan ^{2 }Î¸ cos^{2 }Î¸ =Â 1 âˆ’ cos^{2 }Î¸Â **

**Solution:Â **

We know that,

**sin ^{2Â }Î¸ + cos^{2Â }Î¸ = 1**

Taking,

L.H.S =Â tan^{2Â }Î¸ cos^{2Â }Î¸

= (tan Î¸ Ã— cos Î¸)^{2}

= (sin Î¸)^{2}

= sin^{2 }Î¸

*= 1 – cos ^{2 }Î¸*

= R.H.S

– Hence Proved

**4. cosec Î¸ âˆš(1 â€“ cos ^{2} Î¸) = 1**

**Solution: **

Using identity,

**sin ^{2Â }Î¸ + cos^{2Â }Î¸ = 1Â â‡’ sin^{2Â }Î¸ = 1 – cos^{2Â }Î¸**

Taking L.H.S,

L.H.S = cosec Î¸ âˆš(1 â€“ cos^{2} Î¸)

= cosec Î¸ âˆš( sin^{2Â }Î¸)

= cosec Î¸ x sin^{Â }Î¸

*= 1*

= R.H.S

– Hence Proved

**5. (sec ^{2 }Î¸ âˆ’ 1)(cosec^{2 }Î¸ âˆ’ 1) = 1Â **

**Solution:**

Using identities,

**(sec ^{2 }Î¸ âˆ’ tan^{2 }Î¸) = 1Â and (cosec^{2 }Î¸ âˆ’ cot^{2 }Î¸) = 1**

We have,

L.H.S =Â (sec^{2 }Î¸ – 1)(cosec^{2}Î¸ – 1)

= tan^{2}Î¸ Ã— cot^{2}Î¸

= (tan Î¸ Ã— cot Î¸)^{2}

= (tan Î¸ Ã— 1/tan Î¸)^{2}

= 1^{2}

*= 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¸

= (tan^{2} Î¸ + 1)/ tan Î¸

= sec^{2} Î¸ / tan Î¸ **[âˆµ sec ^{2 }Î¸ âˆ’ tan^{2 }Î¸ = 1]**

= (1/cos^{2} Î¸) x 1/ (sin Î¸/cos Î¸) [âˆµ tan Î¸ = sin Î¸ / cos Î¸]

= cos Î¸/ (sin Î¸ x cos^{2} 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,

**sin ^{2 }Î¸ + cos^{2 }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,

sin^{2 }Î¸ + cos^{2 }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. cos ^{2 }Î¸ + 1/(1 + cot^{2 }Î¸) = 1**

**Solution: **

We already know that,

**cosec ^{2 }Î¸ âˆ’ cot^{2 }Î¸ = 1 and sin^{2 }Î¸ + cos^{2 }Î¸ = 1**

Taking L.H.S,

** **

= cos^{2} A + sin^{2} A

*= 1*

= R.H.S

– Hence Proved

**10. sin ^{2 }A + 1/(1 + tan ^{2 }A) = 1**

**Solution: **

We already know that,

**sec ^{2 }Î¸ âˆ’ tan^{2 }Î¸ = 1Â and sin^{2 }Î¸ + cos^{2 }Î¸ = 1**

Taking L.H.S,

= sin^{2} A + cos^{2} A

*= 1*

= R.H.S

– Hence Proved

**11.**

**Solution: **

We know that,

**sin ^{2 }Î¸ + cos^{2 }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,

**sin ^{2 }Î¸ + cos^{2 }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. tan ^{2 }Î¸ âˆ’ sin^{2 }Î¸Â =Â tan^{2 }Î¸ sin^{2 }Î¸Â **

**Solution: **

Taking L.H.S,

L.H.S = tan^{2 }Î¸ âˆ’ sin^{2 }Î¸**Â **

*= tan ^{2 }Î¸ sin^{2 }Î¸*

= R.H.S

– Hence Proved

**17. (cosecÂ Î¸Â + sinÂ Î¸)(cosecÂ Î¸Â – sinÂ Î¸) =Â cot ^{2}Î¸ + cos^{2}Î¸Â **

**Solution: **

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

On multiplying we get,

= cosec^{2}Â Î¸Â â€“ sin^{2}Â Î¸

=Â (1 + cot^{2}Â Î¸)Â – (1 – cos^{2}Â Î¸) [Using cosec^{2 }Î¸ âˆ’ cot^{2 }Î¸ = 1Â and sin^{2 }Î¸ + cos^{2 }Î¸ = 1]

=Â 1 + cot^{2}Â Î¸Â – 1 + cos^{2}Â Î¸

*= cot ^{2}Â Î¸Â + cos^{2}Â Î¸*

= R.H.S

– Hence Proved

**18. (sec Î¸Â + cosÂ Î¸) (secÂ Î¸Â – cosÂ Î¸) =Â tan ^{2 }Î¸ + sin^{2 }Î¸Â **

**Solution: **

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

On multiplying we get,

= sec^{2}Â Î¸Â â€“ sin^{2}Â Î¸

= (1 + tan^{2}Â Î¸)Â – (1 – sin^{2}Â Î¸) [Using sec^{2 }Î¸ âˆ’ tan^{2 }Î¸ = 1Â and sin^{2 }Î¸ + cos^{2 }Î¸ = 1]

= 1 + tan^{2}Â Î¸Â – 1 + sin^{2}Â Î¸

*=Â tan ^{2 }Î¸Â + sin ^{2}Â 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€“ sin^{2} A / cos^{2} A [After taking L.C.M]

= cos^{2} A / cos^{2} A [âˆµ 1 â€“ sin^{2} A = cos^{2} 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)

** **

= (cos^{2} A/ sin A) (sin^{2} A/ cos A) (1/ sin A cos A) [âˆµ sin^{2 }Î¸ + cos^{2 }Î¸ = 1]

= (sin AÂ cos A)Â (1/ cos A sin A)

*= 1*

= R.H.S

– Hence Proved

**21. (1 +Â tan ^{2 }Î¸)(1 – sinÂ Î¸)(1 + sinÂ Î¸) = 1**

**Solution: **

Taking L.H.S = (1 +Â tan^{2}Î¸)(1 – sin Î¸)(1 + sin Î¸)

And, we know sin^{2}Â Î¸Â + cos^{2}Â Î¸Â = 1 and sec^{2}Â Î¸Â – tan^{2}Â Î¸Â = 1

So,

L.H.S = (1 +Â tan^{2} Î¸)(1 – sinÂ Î¸)(1 + sinÂ Î¸)

= (1 +Â tan^{2 }Î¸){(1 – sinÂ Î¸)(1 + sinÂ Î¸)}

= (1 +Â tan^{2 }Î¸)(1 –Â sin^{2 }Î¸)

= sec^{2 }Î¸ (cos^{2} Î¸)

= (1/ cos^{2} Î¸) x cos^{2} Î¸

*= 1*

= R.H.S

– Hence Proved

**22. sin ^{2 }A cot^{2 }A + cos^{2 }A tan^{2 }AÂ = 1**

**Solution: **

We know that,

cot^{2 }A = cos^{2 }A/ sin^{2} A and tan^{2} A = sin^{2} A/cos^{2} A

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

L.H.S = sin^{2 }A cot^{2 }A + cos^{2 }A tan^{2 }A

= {sin^{2 }A (cos^{2 }A/ sin^{2} A)} + {cos^{2 }A (sin^{2} A/cos^{2} A)}

= cos^{2 }A + sin^{2} A

*= 1 [âˆµ sin ^{2 }Î¸ + cos^{2 }Î¸ = 1]*

= R.H.S

– Hence Proved

**23.**

**Solution: **

(i) Taking the L.H.S and using sin^{2 }Î¸ + cos^{2 }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 sin^{2 }Î¸ + cos^{2 }Î¸ = 1, we have

L.H.S = tan Î¸ â€“ cot Î¸

*= R.H.S*

– Hence Proved

**24. (cos ^{2} Î¸/ sin Î¸) â€“ cosec Î¸ + sin Î¸ = 0**

**Solution: **

Taking L.H.S and using sin^{2 }Î¸ + cos^{2 }Î¸ = 1, we have

= â€“ sin Î¸ + sin Î¸

*= 0*

= R.H.S

– Hence proved

**25.**

**Solution: **

Taking L.H.S,

** **

*= 2 sec ^{2} A*

= R.H.S

– Hence proved

**26. **

**Solution:**

Taking the LHS and using sin^{2 }Î¸ + cos^{2 }Î¸ = 1, we have

= 2/ cos Î¸

*= 2 sec Î¸*

= R.H.S

– Hence proved

**27.**

**Solution: **

Taking the LHS and using sin^{2 }Î¸ + cos^{2 }Î¸ = 1, we have

*= R.H.S*

- Hence proved

**28.**

**Solution: **

Taking L.H.S,

Using sec^{2 }Î¸ âˆ’ tan^{2 }Î¸ = 1Â and cosec^{2 }Î¸ âˆ’ cot^{2 }Î¸ = 1

*= R.H.S*

– Hence proved

**29.**

**Solution: **

Taking L.H.S and using sin^{2 }Î¸ + cos^{2 }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. sec ^{6 }Î¸ = tan^{6 }Î¸ + 3 tan^{2Â }Î¸ sec^{2 }Î¸ + 1**

**Solution:Â **

From trig. Identities we have,

sec^{2 }Î¸ âˆ’ tan^{2 }Î¸ = 1

On cubingÂ both sides,

(sec^{2}Î¸ âˆ’ tan^{2}Î¸)^{3 }= 1

sec^{6 }Î¸ âˆ’ tan^{6 }Î¸ âˆ’ 3sec^{2 }Î¸ tan^{2 }Î¸(sec^{2 }Î¸ âˆ’ tan^{2 }Î¸) = 1

^{3}Â = a

^{3}Â – b

^{3}â€“ 3ab(a – b)]

sec^{6 }Î¸ âˆ’ tan^{6 }Î¸ âˆ’ 3sec^{2 }Î¸ tan^{2 }Î¸ = 1

*â‡’ sec ^{6 }Î¸ = tan^{6 }Î¸ + 3sec^{2 }Î¸ tan^{2 }Î¸ + 1*

L.H.S = R.H.S

– Hence proved

**32. cosec ^{6 }Î¸ = cot^{6 }Î¸ + 3cot^{2 }Î¸ cosec^{2 }Î¸ + 1**

**Solution:**

From trig. Identities we have,

cosec^{2 }Î¸ âˆ’ cot^{2 }Î¸ = 1

On cubingÂ both sides,

(cosec^{2 }Î¸ âˆ’ cot^{2 }Î¸)^{3}Â = 1

cosec^{6 }Î¸ âˆ’ cot^{6 }Î¸ âˆ’ 3cosec^{2 }Î¸ cot^{2 }Î¸ (cosec^{2 }Î¸ âˆ’ cot^{2 }Î¸) = 1

^{3}Â = a

^{3}Â – b

^{3}â€“ 3ab(a – b)]

cosec^{6 }Î¸ âˆ’ cot^{6 }Î¸ âˆ’ 3cosec^{2 }Î¸ cot^{2 }Î¸ = 1

*â‡’ cosec ^{6 }Î¸ = cot^{6 }Î¸ + 3 cosec^{2 }Î¸ cot^{2 }Î¸ + 1*

L.H.S = R.H.S

– Hence proved

**33.**

**Solution: **

Taking L.H.S and using sec^{2 }Î¸ âˆ’ tan^{2 }Î¸ = 1Â â‡’ 1 + tan^{2 }Î¸ = sec^{2 }Î¸

= R.H.S

– Hence proved

**34.**

**Solution: **

Taking L.H.S and using the identity sin^{2}A + cos^{2}AÂ = 1, we get

sin^{2}A = 1 âˆ’ cos^{2}A

â‡’ sin^{2}A = (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 sec^{2 }Î¸ âˆ’ tan^{2 }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 sec ^{2 }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 – cos^{2 }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 – sin^{2 }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 sec^{2 }Î¸ âˆ’ tan^{2 }Î¸ = 1,

**4.**

**Solution: **

Given,

tan Î¸ = 3/4

By using sec^{2 }Î¸ âˆ’ tan^{2 }Î¸ = 1,

** **

sec Î¸ = 5/4

Since, sec Î¸ = 1/ cos Î¸

â‡’ cos Î¸ = 1/ sec Î¸

= 1/ (5/4)

*= 4/5*

So,

**5.Â **

**Solution: **

Given, tan Î¸ = 12/5

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

Now, by using cosec^{2 }Î¸ âˆ’ cot^{2 }Î¸ = 1

cosec Î¸ = âˆš(1 + cot^{2 }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 cosec^{2 }Î¸ âˆ’ cot^{2 }Î¸ = 1, we can find cosec Î¸

cosec Î¸ = âˆš(1 + cot^{2} 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 – sin^{2} 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 cosec^{2 }A âˆ’ cot^{2 }A = 1, we find cot A

= 4/2

*= 2*