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# NCERT Solutions for Class 11 Maths of Chapter 3 Miscellaneous Exercise

The NCERT  maths solutions of chapter 3  miscellaneous exercise containing 10 questions is the extract of the lesson ‘Trigonometry’ . The questions of this exercise are very important for the purpose of your academic exams and entrance exams of engineering.The solutions are explained discretely in order to clear the doubts generally faced by  11 th class students .All questions are explained readily by the teacher expertise in maths. We hope that these solutions by Future Study Point will help you in your preparations of the exam. After you go through each solution write your comment in the form of suggestions, doubts and your requirement.

## NCERT Solutions for Class 11 Maths of Chapter 3 Miscellaneous Exercise

Exercise 3.1 and Exercise 3.2

Exercise 3.3

### CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

Study notes of Maths and Science NCERT and CBSE from class 9 to 12

## Class 11 th CBSE NCERT Maths Solution of Chapter 3 Miscellaneous Exercise

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Q1. Prove that.

$\dpi{100} {\color{DarkBlue} \mathbf{2cos\frac{\Pi }{13}cos\frac{9\Pi }{13}+cos\frac{3\Pi }{13}+cos\frac{5\Pi }{13}=0}}$

L.H.S

$\dpi{100} {\color{DarkBlue} \mathbf{=2cos\frac{\Pi }{13}cos\frac{9\Pi }{13}+cos\frac{3\Pi }{13}+cos\frac{5\Pi }{13}}}$

cosA + cosB =2cos(A+B)/2.cos(A–B)/2

$\dpi{100} {\color{DarkBlue} \mathbf{{ \mathbf{=2cos\frac{\Pi }{13}cos\frac{9\Pi }{13}+2cos\frac{\frac{3\Pi }{13}+\frac{5\Pi }{13}}{2}cos\frac{\frac{3\Pi }{13}-\frac{5\Pi }{13}}{2}}}}}$

$\dpi{100} {\color{DarkBlue} { =\mathbf{2cos\frac{\Pi }{13}cos\frac{9\Pi }{13}+2cos\frac{4\Pi }{13}.cos\left ( -\frac{\Pi }{13} \right )}}}$

As we know cos(–x) = cosx

$\dpi{100} {\color{DarkBlue} { =\mathbf{2cos\frac{\Pi }{13}cos\frac{9\Pi }{13}+2cos\frac{4\Pi }{13}.cos\left ( \frac{\Pi }{13} \right )}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=2cos\frac{\Pi }{13}(cos\frac{9\Pi }{13}+cos\frac{4\Pi }{13})}}$

$\dpi{100} {\color{DarkBlue} =\mathbf{2cos\frac{\Pi }{13}\left ( 2cos\frac{\frac{9\Pi }{13}+\frac{4\Pi }{13}}{2}.cos\frac{\frac{9\Pi }{13}-\frac{4\Pi }{13}}{2} \right )}}$

$\dpi{100} {\color{DarkBlue} =\mathbf{2cos\frac{\Pi }{13}\left ( 2cos\frac{\Pi }{2}.cos\frac{ 5\Pi }{26} \right )}}$

$\dpi{100} {\color{DarkBlue} \mathbf{\because cos\frac{\Pi }{2}=0}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=2cos\frac{\Pi }{13}\times 0\times cos\frac{5\Pi }{26}}}$

= 0 = R.H.S

Q2. Prove that

(sin3x + sinx)sinx  + (cos3x – cosx)cosx = 0

∵sinA + sinB = 2sin(A+B)/2.cos(A–B)/2

∵cosA –cosB = –2sin(A+B)/2.sin(A–B)/2

${\color{DarkBlue} \mathbf{=2sin\frac{3x+x}{2}.cos\frac{3x-x}{2}.sinx-2sin\frac{3x+x}{2}sin\frac{3x-x}{2}.cosx}}$

=2sin2x.cosx.sinx – 2sin2x.sinx.cosx  = 0= R.H.S

Q3.Prove that

${\color{DarkBlue} \mathbf{\left ( cosx+cosy \right )^{2}+\left ( sinx-siny \right )^{2}=4cos^{2}\frac{x+y}{2}}}$

L.H.S

${\color{DarkBlue} =\mathbf{\left ( cosx+cosy \right )^{2}+\left ( sinx-siny \right )^{2}}}$

=cos²x + cos²y + 2cosx.cosy + sin²x + sin²y –2sinx.siny

=(cos²x + sin²x  )+ (cos²y + sin²y) + 2(cosx.cosy – sinx.siny)

=1 + 1 + 2(cosx.cosy – sinx.siny)

=2 + 2(cosx.cosy – sinx.siny)

∵cos(A +B) = cosA.cosB – sinA.sinB

=2 + 2cos(x + y)

=2[1 + cos(x + y)]

∵  cos2A = 2cos²A – 1

⇒1+cos2A  = 2cos²A

$\dpi{100} {\color{DarkBlue} \mathbf{\therefore 1+cos(x+y)=2cos^{2}\left ( \frac{x+y}{2} \right )}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=4cos^{2}\left ( \frac{x-y}{2} \right )}\mathbf{=R.H.S}}$

Q4.Prove that

${\color{DarkGreen} { \mathbf{\left ( cosx -cosy \right )^{2}+\left ( sinx-siny \right )^{2}=4sin^{2}\left ( \frac{x-y}{2} \right )}}}$

L.H.S     (cosx – cosy)² + (sinx – siny)²

=cos²x + cos²y – 2cosx.cosy + sin²x + sin²y –2sinx.siny

=(cos²x  +  sin²x) + (cos²y  + sin²y) –2(cosx.cosy + sinx.siny)

= 1 + 1 – 2cos(x – y)  [∵ cos(A – B) = cosA.cosB + sinA.sinB]

= 2 – 2cos(x– y)

=2[1 – cos(x –y)]

∵ cos2A = 1 – 2sin²A⇒ 1 – cos2A  = 2sin²A

${\color{DarkBlue} \mathbf{\therefore 1- cos\left ( x-y \right )= 2sin^{2}\left ( \frac{x-y}{2} \right )}}$

${\color{DarkBlue} \mathbf{=2\left [ 2sin^{2}\frac{x-y}{2} \right ]}}$

${\color{DarkBlue} \mathbf{=4sin^{2}\left ( \frac{x-y}{2} \right )}}$

= R.H.S

Q5. Prove that  sinx + sin3x + sin5x + sin7x = 4cosx.cos2x.sin4x

L.H.S

=sinx + sin3x + sin5x + sin7x

= (sinx + sin3x )+ (sin5x + sin7x)

∵ sinA + sinB = 2sin(A + B)/2 .cos(A –B)/2

${\color{DarkBlue} \mathbf{=2sin\frac{x+3x}{2}.cos\frac{x-3x}{2}+2sin\frac{5x+7x}{2}.cos\frac{5x-7x}{2}}}$

=2sin2x.cos(–x)  + 2sin6x.cos(–x)

∵ cos(–x) = cosx

=2sin2x.cosx + 2sin6x.cosx

=2cosx(sin2x + sin6x)

∵ sinA + sinB = 2sin(A + B)/2 .cos(A –B)/2

${\color{DarkBlue} \mathbf{=2cosx\left [ 2sin\frac{2x+6x}{2}.cos\frac{2x-6x}{2} \right ]}}$

= 4cosx.sin4x.cos(–2x)

∵ cos(–θ) = cosθ

= 4cosx.cos2x.sin4x = R.H.S

Q6. Prove that

${\color{DarkGreen} \mathbf{\frac{\left ( sin7x+sin5x \right )+\left ( sin9x +sin3x \right )}{\left ( cos7x+cos5x \right )+\left ( cos9x+cos3x \right )}=tan6x}}$

L.H.S

${\color{DarkBlue} ={ \mathbf{\frac{\left ( sin7x+sin5x \right )+\left ( sin9x +sin3x \right )}{\left ( cos7x+cos5x \right )+\left ( cos9x+cos3x \right )}}}}$

${\color{DarkBlue} \mathbf{\because sinA +sinB = 2sin\frac{A+B}{2}.cos\frac{A-B}{2}}}$

${\color{DarkBlue} \mathbf{\because cosA +cosB = 2cos\frac{A+B}{2}.cos\frac{A-B}{2}}}$

$\dpi{120} {\color{DarkBlue} \mathbf{=\frac{\left [ 2sin\left ( \frac{7x+5x}{2}).cos(\frac{7x-5x}{2} \right ) \right ]+\left [ 2sin(\frac{9x+3x}{2}).cos(\frac{9x-3x}{2}) \right ]}{\left [ 2cos\left ( \frac{7x+5x}{2}).cos(\frac{7x-5x}{2} \right ) \right ]+\left [ 2cos\left ( \frac{9x+3x}{2}.cos(\frac{9x-3x}{2} \right ) \right ] }}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{2sin6x.cosx+2sin6x.cos3x}{2cos6x.cosx+2cos6x.cos3x}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{2sin6x(cosx +cos3x)}{2cos6x(cosx +cos3x)}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{2sin6x}{2cos6x}}}$

= tan6x = R.H.S

Q7. Prove that

$\dpi{100} {\color{DarkBlue} \mathbf{sin3x + sin2x-sinx = 4sinx.cos\frac{x}{2}.cos\frac{3x}{2}}}$

Answer. L.H.S = sin3x + (sin2x – sinx)

$\dpi{100} {\color{DarkBlue} \mathbf{\because sinA-sinB = 2cos\frac{A+B}{2}.sin\frac{A-B}{2}}}$

$\dpi{100} {\color{DarkBlue}= \mathbf{sin3x+ 2cos\frac{2x+x}{2}.sin\frac{2x-x}{2}}}$

$\dpi{100} {\color{DarkBlue}= \mathbf{sin3x+ 2cos\frac{3x}{2}.sin\frac{x}{2}}}$

∵sin2A = 2sinA.cosA

$\dpi{100} {\color{DarkBlue} {\therefore \mathbf{sin3x=}\mathbf{2sin\frac{3x}{2}.cos\frac{3x}{2}}}}$

$\dpi{100} {\color{DarkBlue}= \mathbf{2sin\frac{3x}{2}.cos\frac{3x}{2}+ 2cos\frac{3x}{2}.sin\frac{x}{2}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=2cos\frac{3x}{2}\left ( sin\frac{3x}{2}+sin\frac{x}{2} \right )}}$

$\dpi{100} {\color{DarkBlue} \mathbf{\because sinA +sinB = 2sin\frac{A+B}{2}.cos\frac{A-B}{2}}}$

$\dpi{120} {\color{DarkBlue} \mathbf{ = 2cos\frac{3x}{2}.2sin\frac{(\frac{3x}{2}+\frac{x}{2})}{2}.cos\frac{(\frac{3x}{2}-\frac{x}{2})}{2}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=4sinx.cos\frac{x}{2}.cos\frac{3x}{2}= R.H.S}}$

$\dpi{100} {\color{DarkGreen} {\mathbf{Q8.} \mathbf{Find \; sin\frac{x}{2},cos\frac{x}{2}\; and\; \; tan\frac{x}{2},if\; tan\frac{-4}{3},x\; in \; quadrant\; ll.}}}$

Since x is in 2nd quadrant, so the value of x would be estimated as follows

$\dpi{100} {\color{DarkBlue} \mathbf{\frac{\Pi }{2}< x< \Pi }}$

Dividing all the terms of this equation of inequality

$\dpi{100} {\color{DarkBlue} \mathbf{\frac{\Pi }{4}< \frac{x}{2}< \frac{\Pi }{2}}}$

The value of $\dpi{100} {\color{DarkBlue} \mathbf{sin\frac{x}{2},cos\frac{x}{2}\; and \; tan\frac{x}{2}}}$will lye in l quadrant.

We are given that

$\dpi{100} {\color{DarkBlue} \mathbf{tanx =\frac{-4}{3}}}$

$\dpi{100} {\color{DarkBlue} \because \mathbf{tan2x=\frac{2tanx}{1-tan^{2}x}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{\therefore tanx = \frac{2tan\frac{x}{2}}{1-tan^{2}\frac{x}{2}}}}$

Placing the value of tanx =(–4/5)

$\dpi{100} {\color{DarkBlue} \mathbf{\Rightarrow \frac{-4}{3}=\frac{2tan\frac{x}{2}}{1-tan^{2}\frac{x}{2}}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{\Rightarrow 4tan^{2}\frac{x}{2 }-4=6tan\frac{x}{2}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{\Rightarrow 4tan^{2}\frac{x}{2}-6tan\frac{x}{2}-4 =0}}$

$\dpi{100} {\color{DarkBlue} \mathbf{\Rightarrow 4tan^{2}\frac{x}{2}-8tan\frac{x}{2}+tan\frac{x}{2}-4=0}}$

$\dpi{100} {\color{DarkBlue} \mathbf{\Rightarrow 4tan\frac{x}{2}\left ( tan\frac{x}{2}-2 \right )+2\left ( tan\frac{x}{2}-2 \right )=0}}$

$\dpi{100} {\color{DarkBlue} \mathbf{\Rightarrow \left ( tan\frac{x}{2}-2 \right )\left ( 4tan\frac{x}{2}+2 \right )=0}}$

$\dpi{100} {\color{DarkBlue} \mathbf{\Rightarrow tan\frac{x}{2 }=2,tan\frac{x}{2}=-\frac{1}{2}}}$

As we have shown above that x/2 lyes in 1st quadrant,tan(x/2) will be positive

$\dpi{100} {\color{DarkBlue} \mathbf{\therefore tan\frac{x}{2}=2}}$

$\dpi{100} {\color{DarkBlue} \mathbf{sec\frac{x}{2}=\pm \sqrt{1+tan^{2}\frac{x}{2}}=\pm \sqrt{1+2^{2}}=\pm \sqrt{5}}}$

x/2 lies in 1 st quadrant so sec(x/2) =√5

$\dpi{100} {\color{DarkBlue} \mathbf{\therefore cos\frac{x}{2}=\frac{1}{\sqrt{5}}}}$

sinx/2 = ±√(1–cos²x/2)

x/2 lies in 1st quadrant so sin(x/2) will be positive

$\dpi{100} {\color{DarkBlue} \mathbf{sin\frac{x}{2}=\sqrt{1-cos^{2}\frac{x}{2}}=\sqrt{1-\left ( \frac{1}{\sqrt{5}} \right )^{2}}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{sin\frac{x}{2}= \frac{2}{\sqrt{5}}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{Therefore \; \; value\; of\; sin\frac{x}{2},cos\frac{x}{2}\; and\; \; tan\frac{x}{2}\; will\; be\frac{2}{\sqrt{5}},\frac{1}{\sqrt{5}}\; and\; 2.}}$

$\dpi{100} {\color{DarkGreen} {\mathbf{Q9.} \mathbf{Find \; sin\frac{x}{2} , cos\frac{x}{2}\; and\; tan\frac{x}{2}\; for \; cosx=-\frac{1}{3},x\; in\; quadrant\; lll.}}}$

Answer.  x is in lll quadrant, so its value would be estimated as follows

π < x < 3π/2, dividing this eq. by 2,we shall have

π/2 < x/2<3π/4, which reveals that x/2  lyes in ll quadrant .

∵cos2x = 2cos²x –1

$\dpi{100} {\color{DarkBlue} \mathbf{\therefore cosx =2cos^{2}\frac{x}{2}-1}}$

$\dpi{120} {\color{DarkBlue} \mathbf{\Rightarrow cos\frac{x}{2}=\pm \sqrt{\frac{1-\frac{1}{3}}{2}}}}$

$\dpi{100} {\color{DarkBlue} \Rightarrow \mathbf{cos\frac{x}{2}=\pm \frac{1}{\sqrt{3}}}}$

As we have shown above that x/2 lies in ll quadrant and it is known to us ,the value of cos  is negative in 2nd quadrant.

$\dpi{100} {\color{DarkBlue} \therefore \mathbf{cos\frac{x}{2}=-\frac{1}{\sqrt{3}}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{sin\frac{x}{2}=\pm \sqrt{1-cos^{2}\frac{x}{2}}=\pm \sqrt{1-\left ( -\frac{1}{\sqrt{3}} \right )^{2}}=\pm \sqrt{\frac{2}{3}}}}$

As we have shown above that x/2 lies in ll quadrant and it is known to us ,the value of sin x/2 is positive in 2nd quadrant.

$\dpi{100} {\color{DarkBlue} \mathbf{\therefore sin\frac{x}{2}=\sqrt{\frac{2}{3}}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{\because tan\frac{x}{2}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}}}$

$\dpi{120} {\color{DarkBlue} \mathbf{\Rightarrow tan\frac{x}{2}= \frac{\sqrt{\frac{2}{3}}}{-\frac{1}{\sqrt{3}}}}}$

$\dpi{100} {\color{DarkBlue} \Rightarrow \mathbf{tan\frac{x}{2}=-\sqrt{2}}}$

$\dpi{100} {\color{DarkBlue} Therefore\; \mathbf{The \; value \; of\; sin\frac{x}{2},\; cos\frac{x}{2}\; \; and\; \; tan\frac{x}{2} are\; {-\sqrt{\frac{2}{3}}},-\frac{1}{\sqrt{3}}\; and -\sqrt{2}.}}$

$\dpi{100} {\color{DarkBlue} \mathbf{Q10.}\mathbf{Find \; sin\frac{x}{2},cos\frac{x}{2}\; and\; \; tan\frac{x}{2}\; for\; sinx=\frac{1}{4},x\; in \; quadrant\; ll.}}$

Since ,x is in ll quadrant ,its value would be estimated as follows

π/2 < x < π

⇒π/4 < x/2 < π/2, implies that  x/2 lies in l st quadrant, which shows that all sinx/2, cos x/2 and tan x/2 will lye in 1st quadrant.

$\dpi{100} {\color{DarkBlue} \mathbf{cosx=\pm \sqrt{1-sin^{2}x}=\pm \sqrt{1-\left ( \frac{1}{4} \right )^{2}}=\pm \frac{\sqrt{15}}{4}}}$

x is in ll quadrant ,so cosx will be negative

$\dpi{100} {\color{DarkBlue} \therefore \mathbf{cosx =-\frac{\sqrt{15}}{4}}}$

cos2x = 2cos²x – 1

$\dpi{100} {\color{DarkBlue} \mathbf{\therefore cosx = \pm \sqrt{}\frac{1+cos2x}{2}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{\Rightarrow cos\frac{x}{2}=\pm \sqrt{\frac{1+cosx}{2}}}}$

$\dpi{120} {\color{DarkBlue} \mathbf{\Rightarrow cos\frac{x}{2}=\pm \sqrt{\frac{1+\frac{\sqrt{15}}{4}}{2}}}}$

We have shown that x/2 lyes in l quadrant and value of cos is positive in l quadrant.

$\dpi{100} {\color{DarkBlue} \mathbf{\therefore cos\frac{x}{2}=\frac{\sqrt{4+\sqrt{15}}}{2\sqrt{2}}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{sin\frac{x}{2} =\pm \sqrt{1-cos^{2}\frac{x}{2}}}}$

As it is already known to us that x/2 is  in l st quadrant and value of sin x/2 is postitive in lst quadrant.

$\dpi{100} {\color{DarkBlue} \mathbf{\Rightarrow sin\frac{x}{2}=\sqrt{1-\left ( \frac{\sqrt{4-\sqrt{15}}}{2\sqrt{2}} \right )^{2}}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{\Rightarrow sin\frac{x}{2}= \frac{\sqrt{4+\sqrt{15}}}{2\sqrt{2}}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{tan\frac{x}{2} =\frac{sin\frac{x}{2}}{cos\frac{x}{2}}}}$

$\dpi{120} {\color{DarkBlue} \mathbf{tan\frac{x}{2}= \frac{\frac{\sqrt{4+\sqrt{15}}}{2\sqrt{2}}}{\frac{\sqrt{4-\sqrt{15}}}{2\sqrt{2}}}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{tan\frac{x}{2}=\frac{\sqrt{4+\sqrt{15}}}{\sqrt{4-\sqrt{15}}}}}$

Multiplying the denominator and numerator by the conjugate of denominator $\dpi{100} {\color{DarkBlue} \mathbf{\sqrt{4+\sqrt{15}}}}$.

$\dpi{100} {\color{DarkBlue} \mathbf{\Rightarrow \frac{\sqrt{4+\sqrt{15}}}{\sqrt{4-\sqrt{15}}}\times \frac{\sqrt{4+\sqrt{15}}}{\sqrt{4+\sqrt{15}}}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=4 +\sqrt{15}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{The \; value \; of\; sin\frac{x}{2},cos\frac{x}{2}\; and \; \; tan\frac{x}{2}\; are\; \frac{\sqrt{4+\sqrt{15}}}{2\sqrt{2}},\frac{\sqrt{4-\sqrt{15}}}{2\sqrt{2}}\; \; and\; 4+\sqrt{15}.}}$

## NCERT Solutions of Science and Maths for Class 9,10,11 and 12

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### NCERT Solutions for class 11 maths

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