Future Study Point

# NCERT Solutions of Maths for Class 11 Exercise 3.3 Chapter 3-Trigonometric Functions

NCERT Solutions of Maths for Class 11 Exercise 3.3 Chapter 3-Trigonometric Functions are very important for improving maths skills, to solve these questions everybody is required to remember all identities based on trigonometric functions in tips. Here NCERT Solutions of Maths for Class 11 Exercise 3.3 Chapter 3-Trigonometric Functions is presented by Future Study Point which is an associate of Future Point Coaching Center a leading publisher of e-books in Amazon. Solutions of all questions of exercise 3.3 are solved and explained by the maths expert, we are sure every student of class 11 th definitely will like and share it.

## NCERT Solutions of Maths for Class 11 Exercise 3.3 Chapter 3-Trigonometric Functions

Click for online shopping

Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc

Q1.Prove that

$\mathbf{Sin^{2}\frac{\pi }{6}+Cos^{2}\frac{\pi }{3}-tan^{2}\frac{\pi }{4}=-\frac{1}{2}}$

$Sin^{2}\frac{\pi }{6}+Cos^{2}\frac{\pi }{3}-tan^{2}\frac{\pi }{4}$

$\left ( \frac{1}{2} \right )^{2}+\left ( \frac{1}{2} \right )^{2}-(1)^{2}$

$\frac{1}{4}+\frac{1}{4}-1$

$\frac{1}{2}-1=\frac{-1}{2}$

Hence, proved

Q2.Prove that

$\mathbf{2Sin^{2}\frac{\pi }{6}+Cosec^{2}\frac{7\pi }{6}Cos^{2}\frac{\pi }{3}=\frac{3}{2}}$

L.H.S

$2Sin^{2}\frac{\pi }{6}+Cosec^{2}\frac{7\pi }{6}Cos^{2}\frac{\pi }{3}$

$Sin\frac{\pi }{6}=\frac{1}{2},\frac{7\pi }{6}=\pi+\frac{\pi}{6}$

$=2\left ( \frac{1}{2} \right )^{2}+Cosec\left ( \pi+\frac{\pi}{6} \right )\left ( \frac{1}{2} \right )^{2}$

Since, Cosec (π+θ) = – cosecθ

$\therefore Cosec\left ( \pi +\frac{\pi }{6} \right )=-cosec\frac{ \pi}{6}$

$= -\frac{1}{2}+\left ( -cosec\frac{\pi }{6} \right )^{2}.\frac{1}{4}$

$= \frac{1}{2}+(-2)^{2}.\frac{1}{4}$

$= 1+\frac{1}{2}=\frac{3}{2}$

Hence, proved

Q3.Prove that

${\color{DarkGreen} \mathbf{cot^{2}\frac{\Pi }{6}+cosec\frac{5\Pi }{6}+3tan^{2}\frac{\Pi }{6}=6}}$

${\color{DarkBlue} { \mathbf{cot^{2}\frac{\Pi }{6}+cosec\frac{5\Pi }{6}+3tan^{2}\frac{\Pi }{6}=6}}}$

L.H.S

5π/6 = (5×180)/6=150 =180° –30°= π – π/6(π rad.=180°)

$\dpi{100} {\color{DarkBlue} { \mathbf{cot^{2}\frac{\Pi }{6}+cosec\left ( \Pi -\frac{\Pi }{6} \right )+3tan^{2}\frac{\Pi }{6}}}}$

Substituting cotπ/6 =√3, tanπ/6 =1/√3

$\dpi{100} {\color{DarkBlue} \mathbf{\left ( \sqrt{3} \right )^{2}+cocec\frac{\Pi }{6}+3\left ( \frac{1}{\sqrt{3}} \right )^{2}\left [\because cosec\left ( \Pi -\Theta) =cosec\Theta \right ) \right ]}}$

3 + 2 + 1 =6 = R.H.S

Q4. Prove that

${\color{DarkGreen} \mathbf{2sin^{2}\frac{3\Pi }{4}+2cos^{2}\frac{\Pi }{4}+2sec^{2}\frac{\Pi }{3}=10}}$

${\color{DarkBlue} { \mathbf{2sin^{2}\frac{3\Pi }{4}+2cos^{2}\frac{\Pi }{4}+2sec^{2}\frac{\Pi }{3}}}}$

Substituting 3π/4 = (3×180)/4 = 135=180°–45° =π – π/4

$\dpi{100} {\color{DarkBlue} \mathbf{2sin^{2}\left ( \Pi -\frac{\Pi }{4} \right )+2cos^{2}\frac{\Pi }{4}+2sec^{2}\frac{\Pi }{3}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=2sin^{2}\frac{\Pi }{4}+2cos^{2}\frac{\Pi }{4}+2sec^{2}\frac{\Pi }{3}\left [ sin\left ( \Pi -\Theta \right )=sin\Theta \right ]}}$

$\dpi{100} {\color{DarkBlue} \mathbf{\mathbf={2\left ( \frac{1}{\sqrt{2}} \right )^{2}+2\left ( \frac{1}{\sqrt{2}} \right )^{2}+2\times 2^{2}}}}$

= 1 + 1 + 8 =10= R.H.S

Q5. Find  the value of

(i) sin75°

(ii) tan15°

(i) sin75° = sin(45 +30)

sin(45 +30) = sin45.cos30 + cos45.sin30 [sin(x +y) = sinx.cosy + cosx.siny]

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{1}{\sqrt{2}}.\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}.\frac{1}{2}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{\sqrt{3}+1}{2\sqrt{2}}}}$

$\dpi{100} {\color{DarkBlue}\mathbf{Therefore\; sin75^{\circ} \mathbf{\mathbf}{=\frac{\sqrt{3}+1}{2\sqrt{2}}}}}$

(ii) tan15° = tan(45 –30)

$\dpi{100} {\color{DarkBlue} \mathbf{\because tan\left ( A-B \right )=\frac{tanA -tanB}{1+tanA.tanB}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{\because tan\left ( 45-30 \right )=\frac{tan45 -tan30}{1+tan45.tan30}}}$

${\color{DarkBlue} \mathbf{=\frac{1-\frac{1}{\sqrt{3}}}{1+ 1.\frac{1}{\sqrt{3}}}}}$

$\dpi{120} {\color{DarkBlue} \mathbf{= \frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{\sqrt{3}-1}{\sqrt{3}+1}}}$

Multiplying denominator and numerator by (√3 –1)

${\color{DarkBlue} \mathbf{=\frac{\left ( \sqrt{3}-1 \right )^{2}}{\sqrt{3}^{2}-1^{2}}}}$

${\color{DarkBlue} \mathbf{=\frac{\sqrt{3}^{2}+1^{2}-2\sqrt{3}}{3-1}}}$

${\color{DarkBlue} \mathbf{=\frac{4-2\sqrt{3}}{2}}}$

=2– √3

Therefore, tan 15° = 2 – √3

Q6. Prove that

$\dpi{100} {\color{DarkGreen} \mathbf{cos\left ( \frac{\Pi }{4}-x \right )cos\left ( \frac{\Pi }{4}-y \right )-sin\left ( \frac{\Pi }{4}-x \right )sin\left ( \frac{\Pi }{4}-y \right )=sin\left ( x +y \right )}}$

L.H.S

$\dpi{100} {\color{DarkBlue} \mathbf{cos\left ( \frac{\Pi }{4} -x\right )cos\left ( \frac{\Pi }{4}-y \right )-sin\left ( \frac{\Pi }{4}-x \right )sin\left ( \frac{\Pi }{4}-y \right )}}$

Multiplying   and dividing R.H.S by 2.

$\dpi{100} {\color{DarkBlue} =\mathbf{\frac{1}{2}\left [ 2cos\left ( \frac{\Pi }{4}-x \right )cos\left ( \frac{\Pi }{4}-y \right )-2sin\left ( \frac{\Pi }{4} -x\right )sin\left ( \frac{\Pi }{4}-y \right ) \right ]}}$

Applying the identities, 2cosA.cosB = cos(A +B) +cos(A –B)

and –2sinA.sinB = cos(A+B) – cos(A–B)

$\dpi{100} {\color{DarkBlue} =\mathbf{\frac{1}{2}\left [cos\left ( \frac{\Pi }{4}-x +\frac{\Pi }{4}-y \right )+cos\left ( \frac{\Pi }{4}-x-\frac{\Pi }{4} +y\right ) \right ]}}$

$\dpi{100} {\color{DarkBlue} +\mathbf{\frac{1}{2}\left [cos\left ( \frac{\Pi }{4}-x +\frac{\Pi }{4}-y \right )-cos\left ( \frac{\Pi }{4}-x-\frac{\Pi }{4} +y\right ) \right ]}}$

${\color{DarkBlue} \mathbf{=\frac{1}{2}\left [ cos\left \{ \frac{\Pi }{2}-\left ( x+y \right ) \right \}+cos\left \{ \frac{\Pi }{2}-\left ( x-y \right ) \right \} \right ]}}$

${\color{DarkBlue} \mathbf{+\frac{1}{2}\left [ cos\left \{ \frac{\Pi }{2}-\left ( x+y \right ) \right \}-cos\left \{ \frac{\Pi }{2}-\left ( x-y \right ) \right \} \right ]}}$

${\color{DarkBlue} \mathbf{=\frac{1}{2}\left [ sin\left ( x+y \right )+sin\left ( x-y \right ) +sin\left ( x+y \right )-sin\left ( x-y \right )\right ]}}$

${\color{DarkBlue} \mathbf{=\frac{1}{2}\left [ 2sin\left ( x+y \right ) \right ]}}$

= sin(x +y)= R.H.S

Q7. Prove that

$\dpi{100} {\color{DarkGreen} { \mathbf{\frac{tan\left ( \frac{\Pi }{4} +x\right )}{tan\left ( \frac{\Pi }{4}-x \right )}= \left ( \frac{1+tanx}{1-tanx} \right )^{2}}}}$

$\dpi{100} {\color{DarkBlue} { { \mathbf{\frac{tan\left ( \frac{\Pi }{4} +x\right )}{tan\left ( \frac{\Pi }{4}-x \right )}= \left ( \frac{1+tanx}{1-tanx} \right )^{2}}}}}$

L.H.S

$\dpi{100} {\color{DarkBlue} \mathbf{\frac{tan\left ( \frac{\Pi }{4} +x\right )}{tan\left ( \frac{\Pi }{4}-x \right )}}}$

Applying the following identities

$\dpi{100} {\color{DarkBlue} \mathbf{tan\left ( A+B \right ) =\frac{tanA+ tanB}{1-tanA.tanB}\; and\:\; tan\left ( A-B \right )=\frac{tanA-tanB}{1+tanA.tanB}}}$

$\dpi{150} {\color{DarkBlue} \mathbf{= \frac{\frac{tan\frac{\Pi }{4}+tanx}{1-tan\frac{\Pi }{4}.tanx}}{\frac{tan\frac{\Pi }{4}-tanx}{1+tan\frac{\Pi }{4}.tanx}}}}$

$\dpi{150} {\color{DarkBlue} \mathbf{= \frac{\frac{1+tanx}{1-tanx}}{\frac{1-tanx}{1+tanx}}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\left ( \frac{1+tanx}{1-tanx} \right )^{2}}}\mathbf{\color{DarkBlue} \; {= R.H.S}}$

Q8. Prove that

$\dpi{100} {\color{DarkGreen} \mathbf{\frac{cos\left ( \Pi +x \right )cos\left ( -x \right )}{sin\left (\Pi -x \right )cos\left ( \frac{\Pi }{2}+x \right )}=cot^{2}x}}$

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

L.H.S

As we know cos(π+x) = –cosx, cos(–x) =cosx, sin(π–x) =sinx, cos(π/2 +x)=–sinx

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{\left ( -cosx \right )cosx}{\left ( sinx \right )\left ( -sinx \right )}}}$

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

= cot²x = R.H.S

Q9.Prove that

$\dpi{100} {\color{DarkGreen} \mathbf{cos\left ( \frac{3\Pi }{2}+x \right )cos\left ( 2\Pi +x \right )\left [ cot\left ( \frac{3\Pi }{2}+x \right )+cot\left ( 2\Pi +x \right ) \right ]=1}}$

L.H.S

$\dpi{100} {\color{DarkBlue} { \mathbf{cos\left ( \frac{3\Pi }{2}+x \right )cos\left ( 2\Pi +x \right )\left [ cot\left ( \frac{3\Pi }{2}+x \right )+cot\left ( 2\Pi +x \right ) \right ]}}}$

3π/2 +x = 3×180/2+x =270+x=180+90 +x= π + π/2+x

cos(3π/2+x) = cos[π +(π/2 +x)] = –cos(π/2 +x)=–(–sinx)=sinx,cos(2π+x)=cosx

Similarily,cot(3π/2+x)= tanx and cot(2π +x) = cotx

=sinx.cosx(tanx + cotx)

$\dpi{100} {\color{DarkBlue} =\mathbf{sinx.cosx\left ( \frac{sinx}{cosx}+\frac{cosx}{sinx} \right )}}$

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

$\dpi{100} {\color{DarkBlue} \mathbf{=sinx.cosx\left ( \frac{1}{sinx.cosx} \right )}}$

= 1= R.H.S

Q10. Prove that:   sin(n +1)x.sin(n+2)x +cos(n+1)x.cos(n+2)x = cosx

Multiplying and dividing it by 2

$\dpi{100} {\color{DarkBlue} \mathbf{=-\frac{1}{2}\left [ -2 sin(n+1)x.sin(n+2)x\right ]+\frac{1}{2}\left [ 2cos(n+1)x.cos(n+2)x \right ]}}$

$\dpi{100} {\color{DarkBlue} \because \mathbf{-2sinA.sinB=cos\left ( A+B \right )-cos\left ( A-B \right )}}$

$\dpi{100} {\color{DarkBlue} \because \mathbf{2cosA.cosB=cos\left ( A+B \right )+cos\left ( A-B \right )}}$

$\dpi{100} {\color{DarkBlue} \mathbf{-\frac{1}{2}[cos(n+1+n+2)x-cos\left ( n+2-n-1 \right )x]}}$

$\dpi{100} {\color{DarkBlue} +\mathbf{\frac{1}{2}[cos(n+1+n+2)x+cos\left ( n+2-n-1 \right )x]}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{1}{2}\left [ cosx-cos(2n+3) +cosx+ cos(2n+3)x\right ]}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{1}{2}.2cosx}}$

= cosx

Q11. Prove that:

${\color{DarkGreen} { \mathbf{cos\left (\frac{3\Pi }{4}+x \right )-cos\left ( \frac{3\Pi }{4}-x \right )=-\sqrt{2}sinx}}}$

L.H.S

$\dpi{100} {\color{DarkBlue} \mathbf{cos\left ( \frac{3\Pi }{4}+x \right )-cos\left ( \frac{3\Pi }{4}-x \right )}}$

$\dpi{100} {\color{DarkBlue} \because \mathbf{cosA-cosB = -2sin\left ( \frac{A+B}{2} \right ).sin\left ( \frac{A-B}{2} \right )}}$

${\color{DarkBlue} =\mathbf{-2sin\left ( \frac{\frac{3\Pi }{4}+x+\frac{3\Pi }{4}-x}{2} \right ).sin\left ( \frac{\frac{3\Pi }{4}+x-\frac{3\Pi }{4}+x}{2} \right )}}$

${\color{DarkBlue} \mathbf{=-2sin\frac{\frac{6\Pi }{4}}{2}.sinx}}$

${\color{DarkBlue} \mathbf{=-2sin\frac{3\Pi }{4}.sinx}}$

3π/4 = 3 ×180/4 = 135 = 180 –45 = π – π/4

$\dpi{100} {\color{DarkBlue} \mathbf{=-2sin\left ( \Pi -\frac{\Pi }{4} \right ).sinx}}$

sin(π – θ) = sinθ, so

$\dpi{100} {\color{DarkBlue} \mathbf{=-2sin\frac{\Pi }{4}.sinx}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=-2\times \frac{1}{\sqrt{2}}.sinx}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=-\sqrt{2}sinx = R.H.S}}$

Q12.Prove that  sin²6x – sin²4x = sin2x.sin10x

L.H.S

sin²6x – sin²4x

=(sin6x + sin4x)(sin6x  – sin4x)

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

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

$\dpi{100} {\color{DarkBlue} =\mathbf{2sin\frac{6x+4x}{2}.cos\frac{6x-4x}{2}.2cos\frac{6x+4x}{2}sin\frac{6x-4x}{2}}}$

=2sin5x.cosx.2cos5x.sinx

$\dpi{100} {\color{DarkBlue} \mathbf{\because 2sinA.cosA=sin2A}}$

=2sin5x.cos5x. 2sinx.cosx

=sin10x.sin2x=R.H.S

Q13.Prove that

cos²2x – cos²6x = sin4x.sin8x

L.H.S

cos²2x – cos²6x

=(cos2x +cos6x)(cos2x – cos6x)

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

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

$\dpi{100} {\color{DarkBlue} \mathbf{=2cos\frac{2x+6x}{2}.cos\frac{2x-6x}{2}.\left ( -2sin\frac{2x+6x}{2}.sin\frac{2x-6x}{2} \right )}}$

=2cos4x.cos(–2x).[–2sin4x.sin(–2x)]

sin(–θ) = – sinθ, so

= (2sin4x.cos4x).(2sin2x.cos2x)

=sin8x.sin4x = R.H.S

Q14.  Prove that  sin2x + 2sin4x + sin6x = 4cos²x.sin4x

L.H.S

sin2x + 2sin4x + sin6x

(sin2x  + sin6x )+ 2sin4x

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

$\dpi{100} {\color{DarkBlue} \mathbf{ = 2sin\frac{2x+6x}{2}.cos\frac{2x-6x}{2}+2sin4x}}$

= 2sin4x.cos(–2x) + 2sin4x

= 2sin4x.cos2x + 2sin4x

=2sin4x(1  +  cos2x)

As we know  cos2x  = 2cos²x – 1,so

=2sin4x.2cos²x

=4cos²x.sin4x = R.H.S

Q15. Prove that   cot4x(sin5x + sin3x) = cotx(sin5x  – sin3x)

L.H.S

cot4x(sin5x + sin3x)

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

$\dpi{100} {\color{DarkBlue} \mathbf{ = cot4x[2sin\frac{5x+3x}{2}.cos\frac{5x-3x}{2}}}]$

=cot4x[2sin4x.cosx]

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{cos4x}{sin4x}.2sin4x.cosx}}$

=2.cos4x.cosx

For getting cotx in R.H.S ,dividing and multiplying above expression by sinx

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{cosx}{sinx}\left ( 2cos4x.sinx \right )}}$

As we know, 2cosA.sinB = sin(A +B) – sin(A –B)

$\dpi{100} {\color{DarkBlue} \mathbf{=cotx\left [ sin(4x+x)-sin(4x-x) \right ]}}$

=cotx(sin5x  – sin3x)  = R.H.S

Q16. Prove that

$\dpi{100} {\color{DarkGreen} { \mathbf{\frac{sin9x-cos5x}{sin17x-sin3x}=-\frac{sin2x}{cos10x}}}}$

L.H.S

$\dpi{100} {\color{DarkBlue} { { \mathbf{\frac{sin9x-cos5x}{sin17x-sin3x}}}}}$

Applying the following identities

$\dpi{100} {\color{DarkBlue} \mathbf{cosA -cosB =-2sin\frac{A+B}{2}.sin\frac{A-B}{2}}}$

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

$\dpi{120} {\color{DarkBlue} \mathbf{=\frac{-2sin\frac{9x+5x}{2}.sin\frac{9x-5x}{2}}{2cos\frac{17x+3x}{2}.sin\frac{17x-3x}{2}}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{-2sin7x.sin2x}{2cos10x.sin7x}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=-\frac{sin2x}{cos10x}}}{\color{DarkBlue} \mathbf{= R.H.S}}$

Q17. Prove that

$\dpi{100} {\color{DarkGreen} \mathbf{\frac{sin5x+sin3x}{cos5x +cos3x}=tan4x}}$

L.H.S

$\dpi{100} {\color{DarkBlue} { \mathbf{\frac{sin5x+sin3x}{cos5x +cos3x}}}}$

Applying the following identities

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

$\dpi{100} {\color{DarkBlue}{\color{DarkBlue} And\; \; } \mathbf{cosA + cosB = 2cos\frac{A+B}{2}.cos\frac{A-B}{2}}}$

$\dpi{120} {\color{DarkBlue} \mathbf{=\frac{2sin\frac{5x+3x}{2}.cos\frac{5x-3x}{2}}{2cos\frac{5x+3x}{2}.cos\frac{5x-3x}{2}}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{2sin4x.cosx}{2cos4x.cosx}}}$

= tan4x

Q18. Prove that

$\dpi{100} {\color{DarkBlue} \mathbf{\frac{sinx -siny}{cosx +cosy}= tan\frac{x-y}{2}}}$

L.H.S

$\dpi{100} {\color{DarkBlue} \mathbf{\frac{sinx-siny}{cosx +cosy}}}$

Applying the identities

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

$\dpi{100} {\color{DarkBlue} \mathbf{cosA -cosB= 2cos\frac{A+B}{2}.cos\frac{A-B}{2}}}$

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

$\dpi{120} {\color{DarkBlue}= \mathbf{\frac{sin\frac{x-y}{2}}{cos\frac{x-y}{2}}}}$

$\dpi{120} {\color{DarkBlue} \mathbf{=tan\frac{x-y}{2}}}\mathbf{\color{DarkBlue} {= R.H.S}}$

Q19. Prove that

$\dpi{100} {\color{DarkGreen} \mathbf{\frac{sinx+sin3x}{cosx +cos3x}=tan2x}}$

L.H.S

${\color{DarkBlue} \mathbf{\frac{sinx+sin3x}{cosx+cos3x}}}$

Applying the following identities

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

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

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

$\dpi{120} {\color{DarkBlue} \mathbf{=\frac{2sin2x.cos(-x)}{2cos2x.cos(-x)}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{sin2x}{cos2x}}}$

=tan2x

Q20.Prove that

$\dpi{100} {\color{DarkGreen} \mathbf{\frac{sinx-sin3x}{sin^{2}x-cos^{2}x}=2sinx}}$

L.H.S

${\color{DarkBlue} \mathbf{\frac{sinx-sin3x}{sin^{2}x-cos^{2}x}}}$

Applying the  following identities

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

And  cos²x – sin²x = cos2x

$\dpi{120} {\color{DarkBlue} \mathbf{=\frac{2cos\frac{x+3x}{2}.sin\frac{x-3x}{2}}{-cos2x}}}$

${\color{DarkBlue} \mathbf{=\frac{2cos2x.sin(-x)}{-cos2x}}}$

= –2 × –sinx

=2sinx  = R.H.S

Q21.Prove that

${\color{DarkBlue} \mathbf{\frac{cos4x+cos3x+ cos2x}{sin4x+sin3x+sin2x}}}\mathbf{\color{DarkBlue} {=cot3x}}$

L.H.S

${\color{DarkBlue} \mathbf{\frac{cos4x+cos3x+ cos2x}{sin4x+sin3x+sin2x}}}$

Arranging the term as follows

$\dpi{100} {\color{DarkBlue} =\mathbf{\frac{(cos4x+cos2x)+ cos3x}{(sin4x+sin2x)+sin3x}}}$

Applying the following trigonometric identities

cosA +cosB = 2cos(A+B)/2.cos(A–B)/2 and sinA + sinB = 2sin(A+B)/2.cos(A–B)/2

$\dpi{120} {\color{DarkBlue} \mathbf{=\frac{2cos\frac{4x+2x}{2}.cos\frac{4x-2x}{2}+cos3x}{2sin\frac{4x+2x}{2}.cos\frac{4x-2x}{2}+sin3x}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{2cos3x.cosx+cos3x}{2sin3x.cosx+sin3x}}}$

$\dpi{100} {\color{DarkBlue} \mathbf{=\frac{cos3x(2cosx+1)}{sin3x(2cosx+1)}}}$

= cot3x = R.H.S

Q22.Prove that       cotx.cot2x – cot2x.cot3x – cot3x.cotx= 1

Applying the following trigonometric identity

$\dpi{100} {\color{DarkBlue} \mathbf{cot3x=cot(2x+x)= \frac{cot2x.cotx-1}{cotx+cot2x}}}$

cot3x.cotx + cot3x.cot2x = cot2x.cotx – 1

cotx.cot2x –cot2x.cot3x – cot3x.cotx =  1

L.H.S = R.H.S

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

### NCERT Solutions for class 10 maths

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

### NCERT Solutions for class 11 maths

 Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem

CBSE Class 11-Question paper of maths 2015

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

### NCERT solutions for class 12 maths

 Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals

Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2

Class 12 Maths Important Questions-Application of Integrals

Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22

Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22

Solutions of class 12  maths question paper 2021 preboard exam CBSE Solution