Class 11 RD Sharma Solutions – Chapter 5 Trigonometric Functions – Exercise 5.1 | Set 1

Prove the following identities (1 – 13)

Question 1. sec⁴θ – sec²θ = tan⁴θ + tan²θ  

Solution:

We have

sec⁴θ – sec²θ = tan⁴θ + tan²θ

Taking LHS

= sec⁴θ – sec²θ

= sec²θ(sec²θ – 1)

Using sec² θ = tan²θ + 1, we get 

= (1 + tan²θ)tan²θ         

= tan²θ + tan⁴θ

Hence, LHS = RHS (Proved)

Question 2. sin⁶θ + cos⁶θ = 1 – 3sin²θcos²θ      

Solution:

We have

sin⁶θ + cos⁶θ = 1 – 3sin²θcos²θ 

Taking LHS

= sin⁶θ + cos⁶θ   

= (sin²θ)³ + (cos²θ)³

Using a³ + b³ = (a + b)(a² + b² – ab), we get

= (sin²θ + cos²θ)(sin⁴θ + cos⁴θ – sin²θcos²θ)     

Using a² + b² = (a + b)² – 2ab and sin²θ + cos²θ = 1, we get

= (1)[(sin²θ + cos²θ)² – 2sin²θcos²θ – sin²θcos²θ]             

= (1)[(1)² – 3sin²θcos²θ] 

= 1 – 3sin²θcos²θ

Hence, LHS = RHS (Proved)

Question 3. (cosecθ – sinθ)(secθ – cosθ)(tanθ + cotθ) = 1

Solution:

We have

(cosecθ – sinθ)(secθ – cosθ)(tanθ + cotθ) = 1

Taking LHS

= (cosecθ – sinθ)(secθ – cosθ)(tanθ + cotθ)

Using cosecθ = 1/sinθ and secθ = 1/cosθ   

=   

=   

= 

= 1

Hence, LHS = RHS (Proved)

Question 4. cosecθ(secθ – 1) – cotθ(1 – cosθ) = tanθ – sinθ

Solution:

We have

cosecθ(secθ – 1) – cotθ(1 – cosθ) = tanθ – sinθ

Taking LHS

= 

= 

= 

= 

= 

= 

= 

Hence, LHS = RHS(Proved)

Question 5. 

Solution:

We have

Taking LHS

=   

Using a² – b² = (a + b)(a – b) and a³ + b³ = (a + b)(a² + b²ab), we get

= 

= 

= 

= 

= sinA

Hence, LHS = RHS(Proved)

Question 6. 

Solution:

We have

Taking LHS

= 

Using tanA = sinA/cosA and cotA = cosA/sinA, we get

=                             

= 

= 

= 

Using a³ – b³ = (a – b)(a² + b² + ab), we get

=      

= 

= 

Using cosecA = 1/sinA and secA = 1/cosA, we get

= secAcosecA + 1         

Hence, LHS = RHS(Proved)                

Question 7.

Solution:  

We have

Taking LHS

= 

Using a³ ± b³ = (a ± b)(a² + b² ± ab), we get

=  

Using sin²θ + cos²θ = 1, we get

= 1 – sinAcosA + 1 + sinAcosA          

= 2

Hence, LHS = RHS(Proved)

Question 8. (secAsecB + tanAtanB)² – (secAtanB + tanAsecB)² = 1

Solution:

We have

(secAsecB + tanAtanB)² – (secAtanB + tanAsecB)² = 1

Taking LHS

= (secAsecB + tanAtanB)² – (secAtanB + tanAsecB)²

Expanding the above equation using the formula  

(a + b)² = a² + b² + 2ab

= (secAsecB)² + (tanAtanB)² + 2(secAsecB)(tanAtanB) – 

   (secAtanB)² – (tanAsecB)² – 2(secAtanB)(tanAsecB)

= sec²Asec²B + tan²Atan²B – sec²Atan²B – tan²Asec²B

= sec²A(sec²B – tan²B) – tan²A(sec²B – tan²B)

= sec²A – tan²A                -(Using sec²θ – tan²θ = 1)

= 1

Hence, LHS = RHS(Proved)

Question 9. 

Solution:

We have

Taking RHS

= 

= 

= ×

= 

= 

= 

= 

= 

= 

= 

= 

= 

= ×

= 

= 

= 

= 

Hence, RHS = LHS(Proved)

Question 10. 

Solution:

We have

Taking LHS

= 

Using 1 + tan²x = sec²x and 1 + cot²x = cosec²x, we get

= 

= 

= 

= 

= 

Using a² + b² = (a + b)² – 2ab, we get

= 

= 

= 

Hence, LHS = RHS (Proved)

Question 11. 

Solution:

We have

Taking LHS

= 

By using the formulas cotθ = cosθ/sinθ and tanθ = sinθ/cosθ, we get

= 

= 

= 

Using a³+b³ = (a + b)(a² + b² – ab), we get  

=                                   

= 

= 1 – (sin²θ + cos²θ) + sinθcosθ

= 1 – 1 + sinθcosθ

= sinθcosθ

Hence, LHS = RHS (Proved)

Question 12.

Solution:

We have

= 

Taking LHS

= 

=      

= 

= 

= 

= 

= 

= 

= 

Hence, LHS = RHS(Proved)

Question 13. (1 + tanαtanβ)² + (tanα – tanβ)²  = sec²αsec²β

Solution:

We have

(1 + tanαtanβ)² + (tanα – tanβ)² = sec^2αsec²β

Taking LHS

= (1 + tanαtanβ)² + (tanα – tanβ)²

= (1 + tan²αtan²β + 2tanαtanβ) + (tan²α + tan²β – 2tanαtanβ)

= 1 + tan²αtan²β + tan²α + tan²β

= (1 + tan²β) + tan²α(1 + tan²β)

= (1 + tan²β)(1 + tan²α)

= sec²αsec²β

Hence, LHS = RHS (Proved)