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Class 11 RD Sharma Solutions – Chapter 9 Trigonometric Ratios of Multiple and Submultiple Angles – Exercise 9.1 | Set 1
  • Last Updated : 08 May, 2021

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 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. \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 knwo 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.




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