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

Prove the following identities: 

Question 1. √[(1 - cos2x)/(1 + cos2x)] = tanx

Solution:

Let us solve LHS,

= √[(1 - cos2x)/(1 + cos2x)]

As we know that, cos2x =1 - 2 sin²x

= 2 cos²x - 1

So,

= √[(1 - cos2x)/(1 + cos2x)] 

= √[(1 - (1 - 2sin²x))/(1 + (2cos²x - 1))]

= √(1 - 1 + 2sin²x)/(1 + 2cos²x - 1)1

= √[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 sin²x 

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. \sqrt{2+\sqrt{2+cos4x}}=2cosx , 0 < r < π/4

Solution:

Let us solve LHS,

\sqrt{2+\sqrt{2+cos4x}}=\sqrt{2+\sqrt{2+2(2cos^22x-1)}}

As we know that,

 cos 2x = 2 cos²x - 1 ⇒ cos 4x = 2 cos²2x - 1

So, 

= \sqrt{2+\sqrt{2+4cos^22x-2}}

= \sqrt{2+\sqrt{4cos^22x}}

= \sqrt{2+2cos2x}

= \sqrt{2+\sqrt{2(2cos^2x-1)}}

= \sqrt{2+4cos^2x-2}

= \sqrt{4cos^2x}

= 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 − 1 + 2sin²x + 2sinxcosx} / {1 + 2cos²x − 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 − 1)}

= {sinx + 2 sinx cos x} / {1 + cosx + 2cos²x − 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 (π/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 – 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/√2, we get,

= \frac{\frac{1}{√2}(cosx-sinx)}{\frac{1}{√2}(sinx+cosx)}

= \frac{(\frac{1}{√2}.cosx-\frac{1}{√2}.sinx)}{(\frac{1}{√2}.sinx+\frac{1}{√2}.cosx)}

= \frac{sin(\frac{π}{4})cosx-cos(\frac{π}{4})sinx}{sin(\frac{π}{4})sinx+cos(\frac{π}{4})cosx}

Since, 1/√2 = sin π/4, so

= \frac{sin(\frac{π}{4}-x)}{cos(\frac{π}{4}-x)}

By using the formulas, we get

sin(A - B) = sinA cosB - sinB cosA 

cos(A - B)= cosA cosB + sinA sinB

= tan (π/4 - x)

LHS = RHS

Hence proved.

Question 8. cos x/(1 - sin x) = tan (π/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,

\frac{cosx}{1-sinx}=\frac{cos^2(\frac{x}{2})-sin^2(\frac{x}{2})}{1-2sin(\frac{x}{2})cos(\frac{x}{2})}

= \frac{[cos(\frac{x}{2})-sin(\frac{x}{2})][cos(\frac{x}{2})+sin(\frac{x}{2})]}{sin^2(\frac{x}{2})+cos^2(\frac{x}{2})+2sin(\frac{x}{2})cos(\frac{x}{2})}

By using the formulas, 

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

= \frac{[cos(\frac{x}{2})-sin(\frac{x}{2})][cos(\frac{x}{2})+sin(\frac{x}{2})]}{[sin(\frac{x}{2})+cos(\frac{x}{2})]^2}

= \frac{[cos(\frac{x}{2})-sin(\frac{x}{2})][cos(\frac{x}{2})+sin(\frac{x}{2})]}{[sin(\frac{x}{2})+cos(\frac{x}{2})][sin(\frac{x}{2})+cos(\frac{x}{2})]}

= \frac{cos(\frac{x}{2})-sin(\frac{x}{2})}{sin(\frac{x}{2})+cos(\frac{x}{2})}

= \frac{cos(\frac{x}{2})+sin(\frac{x}{2})}{sin(\frac{x}{2})-cos(\frac{x}{2})}

Now multiply numerator and denominator by 1/√2, we get,

= \frac{\frac{1}{√2}[cos(\frac{x}{2})+sin(\frac{x}{2})]}{\frac{1}{√2}[sin(\frac{x}{2})-cos(\frac{x}{2})]}

= \frac{(\frac{1}{√2})cos(\frac{x}{2})+(\frac{1}{√2})sin(\frac{x}{2})}{(\frac{1}{√2})sin(\frac{x}{2})-(\frac{1}{√2})cos(\frac{x}{2})}

= \frac{sin(\frac{π}{4})cos(\frac{x}{2})+cos(\frac{π}{4})sin(\frac{x}{2})}{sin(\frac{π}{4})sin(\frac{x}{2})-cos(\frac{π}{4})cos(\frac{x}{2})}

= \frac{sin(π/4-x)}{cos(π/4-x)}

 = tan (π/4 - x)

LHS = RHS

Hence proved.

Question 9. cos² π/8 + cos² 3π/8 + cos² 5π/8 + cos² 7π/8 = 2

Solution:

Let us solve LHS,

= cos² π/8 + cos² 3π/8 + cos² 5π/8 + cos² 7π/8

As we know that,

cos 2x = 2cos²x - 1

cos 2x+1=2cos² x

cos²x = (Cos 2x + 1)/2

So,

= cos² π/8 + cos² 3π/8 + cos² 5π/8 + cos² 7π/8

= \frac{1+cos(\frac{2π}{8})}{2}+\frac{1+cos(\frac{6π}{8})}{2}+\frac{1+cos(\frac{10π}{8})}{2}+\frac{1+cos(\frac{14π}{8})}{2}

= \frac{1+cos(\frac{2π}{8})}{2}+\frac{1+cos(π-\frac{2π}{8})}{2}+\frac{1+cos(π+\frac{2π}{8})}{2}+\frac{1+cos(2π-\frac{2π}{8})}{2}

= \frac{1+cos(\frac{2π}{8})}{2}+\frac{1-cos(\frac{2π}{8})}{2}+\frac{1-cos(\frac{2π}{8})}{2}+\frac{1+cos(\frac{2π}{8})}{2}

As we know that, cos (π - A) =- cos A, cos (π+ A) = -cos A and cos (2π - A) = cos A

= 2 x {1 + cos(2π/8)/2} + 2 x {1 - cos(2π/8)/2}

= 1 + cos(2π/8) + 1 - cos(2π/8)

= 2 

LHS = RHS

Hence Proved.

Question 10. sin² π/8 + sin² 3π/8 + sin² 5π/8 + sin² 7π/8

Solution:

Let us solve LHS,

= sin² π/8 + sin² 3π/8 + sin² 5π/8 + sin² 7π/8

As we know that, 

cos 2x = 1 - 2sin²x

2sin²x = 1 - cos 2x

sin²x = (1 - cos 2x)/2

So,

= \frac{1-cos(\frac{2π}{8})}{2}+\frac{1-cos(\frac{6π}{8})}{2}+\frac{1-cos(\frac{10π}{8})}{2}+\frac{1-cos(\frac{14π}{8})}{2}

= \frac{1-cos(\frac{2π}{8})}{2}+\frac{1-cos(π-\frac{2π}{8})}{2}+\frac{1-cos(π+\frac{2π}{8})}{2}+\frac{1-cos(2π-\frac{2π}{8})}{2}

= \frac{1-cos(\frac{2π}{8})}{2}+\frac{1+cos(\frac{2π}{8})}{2}+\frac{1+cos(\frac{2π}{8})}{2}+\frac{1-cos(\frac{2π}{8})}{2}

As we know, cos (π - A) = -cos A, cos (π + A) = -cos A and cos (2π - A) = cos A

= 2 x {1 - cos(2π/8)/2} + 2 x {1 + cos(2π/8)/2}

= 1 - cos(2π/8) + 1 + cos(2π/8)

= 2

LHS = RHS

Hence proved.

Question 11. (cos α + cos β)² + (sin α + sin β)² = 4 cos²(α - β)/2

Solution:

Let us solve LHS,

= (cos α+ cos β)²+ (sin α+ sin β)²

On expanding, we get,

= cos²α + cos² β + 2 cos α cos β + sin²α+ sin²β + 2 sin α sin β

= 2+2 cos α cos β + 2 sin α sin β

= 2 (1+ cos α cos β+ sin α sin β)

= 2 (1 + cos (α - β))                        [Using, cos (A - B) = cos A cos B+ sin A sin B]

= 2 (1 + 2 cos²(α - β)/2 - 1)            [Using, cos2x = 2cos²x - 1]

= 2 (2 cos²(α - β)/2)

= 4 cos²(α - β)/2

LHS = RHS

Hence Proved.

Question 12. sin²(π/8 + x/2) - sin²(π/8 - x/2) = 1/√2 sin x

Solution:

Let us solve LHS,

= sin²(π/8 + x/2) - sin²(π/8 - x/2)

As we know that, 

sin²A - sin²B = sin (A + B) sin (A-B)

So,

sin²(π/8 + x/2) - sin²(π/8 - x/2) = sin (π/8 + x/2 + π/8 - x/2) sin (π/8 + x/2 - (π/8 - x/2))

= sin (π/8 + π/8) sin (π/8 + x/2 - π/8 + x/2)

= sin π/4 sin x

= 1/√2 sin x                       [As we know that, π/4 = 1/√2]

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 cos³ 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.