# Class 11 RD Sharma Solutions – Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles – Exercise 9.1 | Set 1

### Question 1. âˆš[(1 – cos2x)/(1 + cos2x)] = tanx

Solution:

Let us solve LHS,

= âˆš[(1 – cos2x)/(1 + cos2x)]

As we know that, cos2x =1 – 2 sin2x

= 2 cos2x – 1

So,

= âˆš[(1 – cos2x)/(1 + cos2x)]

= âˆš[(1 – (1 – 2sin2x))/(1 + (2cos2x – 1))]

= âˆš(1 – 1 + 2sin2x)/(1 + 2cos2x – 1)1

= âˆš[2 sin2x/2 cos2x]

= sinx / cosx

= tanx

LHS = RHS

Hence proved.

### Question 2. sin2x/(1 – cos2x) = cotx

Solution:

Let us solve LHS,

= sin 2x/(1 – cos 2x)

As we know that,

cos 2x = 1 – 2 sin2

Sin 2x = 2 sin x cos x

So,

sin 2x/(1-cos 2x) = (2 sin x cos x)/(1 – (1 – 2sin2x))

= (2 sin x cos x)/(1 – 1 + 2sin2x)]

= [2 sin x cos x/2 sin2x]

= cos x/sin x

= cot x

LHS = RHS

Hence proved.

### Question 3. sin 2x/(1 + cos 2x) = tan x

Solution:

Let us solve LHS,

= sin 2x / (1+cos 2x)

As we know that,

cos 2x = 1 – 2 sin2x

= 2 cos2x – 1

sin 2x = 2 sin x cos x

So,

sin 2x / (1 + cos2x) = [2 sin x cos x / (1 + (2cos2x – 1))]

= [2 sin x cos x / (1+2cos2x – 1)]

= [2 sin x cos x/2 cos2x]

= sin x/cos x

= tan x

LHS = RHS

Hence proved.

### Question 4. , 0 < r < Ï€/4

Solution:

Let us solve LHS,

As we know that,

cos 2x = 2 cos2x – 1 â‡’ cos 4x = 2 cos22x – 1

So,

= 2 cos x

LHS = RHS

Hence proved.

### Question 5. [1 – cos 2x + sin 2x]/[1 + cos 2x + sin 2x] = tan x

Solution:

Let us solve LHS,

= [1 – cos 2x + sin 2x]/[1 + cos 2x + sin 2x]

As we know that,

cos 2x = 1 – 2 sin2x

= 2 cos2x – 1

sin 2x = 2 sin x cos x

So,

= {1 – (1 – 2sin2x) + 2sinxcosx} / {1 + (2 cos2x – 1) + 2 sin x cosx}

= {1 âˆ’ 1 + 2sin2x + 2sinxcosx} / {1 + 2cos2x âˆ’ 1 + 2sinx cosx}

= {2 sin2x + 2sinxcosx} / {2 cos2x + 2sinxcosx}

= {2sinx (sinx + cosx)} / {2 cos x (cosx + sin x)}

= sinx/cosx

= tan x

LHS = RHS

Hence proved.

### Question 6. [sin x + sin 2x]/[1 + cos x + cos2x] = tanx

Solution:

Let us solve LHS,

= [sin x + sin 2x]/[1 + cos x + cos 2x]

As we know that,

cos 2x = cos2x sin2x

sin 2x = 2 sin x cos x

So,

{sin x + sin 2x} / {1 + cos x + cos 2x} = {sin x + 2 sin x cos x} / {1 + cosx + (2cos2x âˆ’ 1)}

= {sinx + 2 sinx cos x} / {1 + cosx + 2cos2x âˆ’ 1}

= {sin x + 2 sin x cosx} / {cosx + 2cos2x}

= {sinx (1 + 2 cos x)} / {cosx(1 + 2cosx)}

= sinx / cosx

= tan x

LHS = RHS

Hence proved.

### Question 7. cos 2x / (1+ sin 2x) = tan (Ï€/4 – x)

Solution:

Let us solve LHS,

= cos 2x / (1 + sin 2x)

As we know that,

cos 2x = cos2x – sin2x

sin 2x = 2 sin x cos x

So,

{cos 2x} / {1 + sin 2x} = {cos2x – sin2x} / {1 + 2 sin x cos x}

= {(cosx â€“ sinx)(cosx + sinx)} / {sin2x + cos2x + 2 sin x cos x}

Since, a2 – b2 = (a – b)(a + b) and sin2x + cos2x = 1

So,

= {(cosx – sinx)(cosx + sinx)} / {(sinx + cos x)2

Since, a2+ b2 + 2ab = (a + b)2

So,

= {(cosx – sinx)(cosx + sinx)} / {(sinx + cosx)(sinx + cosx)}

= (cosx – sinx) / (sin x + cos x)

Now multiplying numerator and denominator by 1/âˆš2, we get,

Since, 1/âˆš2 = sin Ï€/4, so

By using the formulas, we get

sin(A – B) = sinA cosB – sinB cosA

cos(A – B)= cosA cosB + sinA sinB

= tan (Ï€/4 – x)

LHS = RHS

Hence proved.

### Question 8. cos x/(1 – sin x) = tan (Ï€/4 + x/2)

Solution:

Let us solve LHS,

= cos x/(1 – sin x)

As we know that,

cos 2x = cos2x – sin2x

cos x = cos2 x/2 – sin2 x/2

sin 2x = 2 sin x cos x

sin x = 2 sin x/2 cos x/2

So,

By using the formulas,

a2 – b2 = (a – b)(a + b) and sin2x + cos2x = 1), we get

Now multiply numerator and denominator by 1/âˆš2, we get,

= tan (Ï€/4 – x)

LHS = RHS

Hence proved.

### Question 9. cos2 Ï€/8 + cos2 3Ï€/8 + cos2 5Ï€/8 + cos2 7Ï€/8 = 2

Solution:

Let us solve LHS,

= cos2 Ï€/8 + cos2 3Ï€/8 + cos2 5Ï€/8 + cos2 7Ï€/8

As we know that,

cos 2x = 2cos2x – 1

cos 2x+1=2cos2 x

cos2x = (Cos 2x + 1)/2

So,

= cos2 Ï€/8 + cos2 3Ï€/8 + cos2 5Ï€/8 + cos2 7Ï€/8

As we know that, cos (Ï€ – A) =- cos A, cos (Ï€+ A) = -cos A and cos (2Ï€ – A) = cos A

= 2 x {1 + cos(2Ï€/8)/2} + 2 x {1 – cos(2Ï€/8)/2}

= 1 + cos(2Ï€/8) + 1 – cos(2Ï€/8)

= 2

LHS = RHS

Hence Proved.

### Question 10. sin2 Ï€/8 + sin2 3Ï€/8 + sin2 5Ï€/8 + sin2 7Ï€/8

Solution:

Let us solve LHS,

= sin2 Ï€/8 + sin2 3Ï€/8 + sin2 5Ï€/8 + sin2 7Ï€/8

As we know that,

cos 2x = 1 – 2sin2x

2sin2x = 1 – cos 2x

sin2x = (1 – cos 2x)/2

So,

As we know, cos (Ï€ – A) = -cos A, cos (Ï€ + A) = -cos A and cos (2Ï€ – A) = cos A

= 2 x {1 – cos(2Ï€/8)/2} + 2 x {1 + cos(2Ï€/8)/2}

= 1 – cos(2Ï€/8) + 1 + cos(2Ï€/8)

= 2

LHS = RHS

Hence proved.

### Question 11. (cos Î± + cos Î²)2 + (sin Î± + sin Î²)2 = 4 cos2(Î± – Î²)/2

Solution:

Let us solve LHS,

= (cos Î±+ cos Î²)2+ (sin Î±+ sin Î²)2

On expanding, we get,

= cos2Î± + cos2 Î² + 2 cos Î± cos Î² + sin2Î±+ sin2Î² + 2 sin Î± sin Î²

= 2+2 cos Î± cos Î² + 2 sin Î± sin Î²

= 2 (1+ cos Î± cos Î²+ sin Î± sin Î²)

= 2 (1 + cos (Î± – Î²))                        [Using, cos (A – B) = cos A cos B+ sin A sin B]

= 2 (1 + 2 cos2(Î± – Î²)/2 – 1)            [Using, cos2x = 2cos2x – 1]

= 2 (2 cos2(Î± – Î²)/2)

= 4 cos2(Î± – Î²)/2

LHS = RHS

Hence Proved.

### Question 12. sin2(Ï€/8 + x/2) – sin2(Ï€/8 – x/2) = 1/âˆš2 sin x

Solution:

Let us solve LHS,

= sin2(Ï€/8 + x/2) – sin2(Ï€/8 – x/2)

As we know that,

sin2A – sin2B = sin (A + B) sin (A-B)

So,

sin2(Ï€/8 + x/2) – sin2(Ï€/8 – x/2) = sin (Ï€/8 + x/2 + Ï€/8 – x/2) sin (Ï€/8 + x/2 – (Ï€/8 – x/2))

= sin (Ï€/8 + Ï€/8) sin (Ï€/8 + x/2 – Ï€/8 + x/2)

= sin Ï€/4 sin x

= 1/âˆš2 sin x                       [As we know that, Ï€/4 = 1/âˆš2]

LHS = RHS

Hence proved.

### Question 13. 1 + cos22x = 2 (cos4x + sin4x)

Solution:

Let us solve LHS,

= 1 + cos22x

As we know that,

cos2x = cos2x sin2x

cos2x+ sin2x = 1

So,

1 + cos22x = (cos2x+ sin2x)2 + (cos2x – sin2x)2

= (cos4x + sin4x + 2 cos2x sin2x) + (cos4x + sin4x – 2cos2x sin2x)

= cos4x + sin4x + cos4x + sin4x

= 2 cos4x + 2 sin4x

= 2 (cos4x + sin4x)

LHS = RHS

Hence proved.

### Question 14. cos32x + 3 cos 2x = 4 (cos6x – sin6x)

Solution:

Let us solve RHS,

= 4 (cos6 x – sin6 x)

On expanding, we get,

4 (cos6 x – sin6 x) = 4 [(cos2x)3 – (sin2x)3]

= 4 (cos2x – sin2x) (cos4x + sin4x + cos2x sin2x)

Now, using the formula, we get

a3 – b3 = (a – b) (a2 + b2 + ab)

= 4 cos 2x (cos4x + sin4x + cos2x sin2x + cos2x sin2x – cos2x sin2x

As we know that,

cos 2x = cos2x – sin2x

So,

= 4 cos 2x (cos4x + sin4x + 2 cos2x sin2x – cos2x sin2x)

= 4 cos 2x [(cos2x)2 + (sin2x)2 + 2 cos2x sin2x – cos2x sin2x]

By using formula,

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

= 4 cos 2x [(1)2 – 1/4 (4 cos2 x sin2 x)]

= 4 cos 2x [(1)2-1/4 (2 cos x sin x)2

Since

sin 2x = 2sin x cos x

= 4 cos 2x [(12) – 1/4 (sin 2x)2]

= 4 cos 2x (1 – 1/4 sin2 2x)

Since

sin2 x = 1 – cos2x

= 4 cos 2x [1 – 1/4 (1 – cos2 2x)]

= 4 cos 2x [1 – 1/4 + 1/4 cos2 2x]

= 4 cos 2x [3/4 + 1/4 cos2 2x]

= 4 (3/4 cos 2x + 1/4 cosÂ³ 2x)

= 3 cos 2x + cos3 2x

= cos3 2x + 3 cos 2x

LHS = RHS

Hence proved.

### Question 15. (sin 3A + sin A) sin A + (cos 3A – cos A) cos A = 0

Solution:

Let us solve LHS,

= (sin 3A + sin A) sin A + (cos 3A – cos A) cos A

= (sin 3A) (sin A) + sin2A + (cos 3A) (cos A) – cos2A

= [(sin 3A) (sin A) + (cos 3A) (cos A)] + (sin2A – cos2A)

= [(sin 3A) (sin A) + (cos 3A) (cos A)] – (cos2A – sin2A)

= cos (3A – A) – cos 2A

As we know that,

cos 2x = cos2A – sin2A

cos A cos B + sin A sin B = cos(A – B)

So,

= cos 2A – cos 2A

= 0

LHS = RHS

Hence Proved.

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