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

### Question 1. sec4Î¸ – sec2Î¸ = tan4Î¸ + tan2Î¸

Solution:

We have

sec4Î¸ – sec2Î¸ = tan4Î¸ + tan2Î¸

Taking LHS

= sec4Î¸ – sec2Î¸

= sec2Î¸(sec2Î¸ – 1)

Using sec2 Î¸ = tan2Î¸ + 1, we get

= (1 + tan2Î¸)tan2Î¸

= tan2Î¸ + tan4Î¸

Hence, LHS = RHS (Proved)

### Question 2. sin6Î¸ + cos6Î¸ = 1 – 3sin2Î¸cos2Î¸

Solution:

We have

sin6Î¸ + cos6Î¸ = 1 – 3sin2Î¸cos2Î¸

Taking LHS

= sin6Î¸ + cos6Î¸

= (sin2Î¸)3 + (cos2Î¸)3

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

= (sin2Î¸ + cos2Î¸)(sin4Î¸ + cos4Î¸ – sin2Î¸cos2Î¸)

Using a2 + b2 = (a + b)2 – 2ab and sin2Î¸ + cos2Î¸ = 1, we get

= (1)[(sin2Î¸ + cos2Î¸)2 – 2sin2Î¸cos2Î¸ – sin2Î¸cos2Î¸]

= (1)[(1)2 – 3sin2Î¸cos2Î¸]

= 1 – 3sin2Î¸cos2Î¸

Hence, LHS = RHS (Proved)

### Question 3. (cosecÎ¸ – sinÎ¸)(secÎ¸ – cosÎ¸)(tanÎ¸ + cotÎ¸) = 1

Solution:

We have

(cosecÎ¸ – sinÎ¸)(secÎ¸ – cosÎ¸)(tanÎ¸ + cotÎ¸) = 1

Taking LHS

= (cosecÎ¸ – sinÎ¸)(secÎ¸ – cosÎ¸)(tanÎ¸ + cotÎ¸)

Using cosecÎ¸ = 1/sinÎ¸ and secÎ¸ = 1/cosÎ¸

= 1

Hence, LHS = RHS (Proved)

### Question 4. cosecÎ¸(secÎ¸ – 1) – cotÎ¸(1 – cosÎ¸) = tanÎ¸ – sinÎ¸

Solution:

We have

cosecÎ¸(secÎ¸ – 1) – cotÎ¸(1 – cosÎ¸) = tanÎ¸ – sinÎ¸

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

Using sin2Î¸ + cos2Î¸ = 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 sec2Î¸ – tan2Î¸ = 1)

= 1

Hence, LHS = RHS(Proved)

### Question 9.

Solution:

We have

Taking RHS

Ă—

Ă—

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 cotÎ¸ = cosÎ¸/sinÎ¸ and tanÎ¸ = sinÎ¸/cosÎ¸, we get

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

= 1 – (sin2Î¸ + cos2Î¸) + sinÎ¸cosÎ¸

= 1 – 1 + sinÎ¸cosÎ¸

= sinÎ¸cosÎ¸

Hence, LHS = RHS (Proved)

### Question 12.

Solution:

We have

Taking LHS

Hence, LHS = RHS(Proved)

### Question 13. (1 + tanÎ±tanÎ˛)2 + (tanÎ± – tanÎ˛)2  = sec2Î±sec2Î˛

Solution:

We have

(1 + tanÎ±tanÎ˛)2 + (tanÎ± – tanÎ˛)2 = sec^2Î±sec2Î˛

Taking LHS

= (1 + tanÎ±tanÎ˛)2 + (tanÎ± – tanÎ˛)2

= (1 + tan2Î±tan2Î˛ + 2tanÎ±tanÎ˛) + (tan2Î± + tan2Î˛ – 2tanÎ±tanÎ˛)

= 1 + tan2Î±tan2Î˛ + tan2Î± + tan2Î˛

= (1 + tan2Î˛) + tan2Î±(1 + tan2Î˛)

= (1 + tan2Î˛)(1 + tan2Î±)

= sec2Î±sec2Î˛

Hence, LHS = RHS (Proved)

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