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

• Last Updated : 21 Feb, 2021

### Question 1. sec4θ – sec2θ = tan4θ + tan2θ

Solution:

We have

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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)

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