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

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Q1.Prove that

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

Answer. L.H.S

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

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

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

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

Hence, proved

Q2.Prove that

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

Answer.

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

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

Answer.

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

L.H.S

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

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

Substituting cotπ/6 =√3, tanπ/6 =1/√3

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

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

Q4. Prove that

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

Answer. L.H.S

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

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

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

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

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

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

Q5. Find the value of

(i) sin75°

(ii) tan15°

Answer.

(i) sin75° = sin(45 +30)

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

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

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

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

(ii) tan15° = tan(45 –30)

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

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

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

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

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

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

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

Answer.

L.H.S

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

Multiplying and dividing R.H.S by 2.

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

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

and –2sinA.sinB = cos(A+B) – cos(A–B)

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

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

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

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

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

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

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

Q7. Prove that

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

Answer.

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

L.H.S

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

Applying the following identities

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

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

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

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

Q8. Prove that

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

Answer.

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

L.H.S

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

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

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

= cot²x = R.H.S

Q9.Prove that

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

Answer.

L.H.S

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

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)

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

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

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

= 1= R.H.S

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

Answer. L.H.S sin(n +1)x.sin(n+2)x +cos(n+1)x.cos(n+2)x

Multiplying and dividing it by 2

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

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

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

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

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

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

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

= cosx

Q11. Prove that:

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

Answer.

L.H.S

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

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

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

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

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

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

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

sin(π – θ) = sinθ, so

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

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

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

Q12.Prove that sin²6x – sin²4x = sin2x.sin10x

Answer.

L.H.S

sin²6x – sin²4x

=(sin6x + sin4x)(sin6x – sin4x)

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

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

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

=2sin5x.cosx.2cos5x.sinx

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

=2sin5x.cos5x. 2sinx.cosx

=sin10x.sin2x=R.H.S

Q13.Prove that

cos²2x – cos²6x = sin4x.sin8x

Answer

L.H.S

cos²2x – cos²6x

=(cos2x +cos6x)(cos2x – cos6x)

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

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

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

=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

Answer.

L.H.S

sin2x + 2sin4x + sin6x

(sin2x + sin6x )+ 2sin4x

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

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

= 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)

Answer.

L.H.S

cot4x(sin5x + sin3x)

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

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

=cot4x[2sin4x.cosx]

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

=2.cos4x.cosx

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

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

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

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

=cotx(sin5x – sin3x) = R.H.S

Q16. Prove that

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

Answer.

L.H.S

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

Applying the following identities

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

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

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

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

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

Q17. Prove that

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

Answer.

L.H.S

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

Applying the following identities

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

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

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

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

= tan4x

Q18. Prove that

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

Answer.

L.H.S

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

Applying the identities

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

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

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

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

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

Q19. Prove that

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

Answer.

L.H.S

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

Applying the following identities

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

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

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

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

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

=tan2x

Q20.Prove that

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

Answer.

L.H.S

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

Applying the following identities

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

And cos²x – sin²x = cos2x

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

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

= –2 × –sinx

=2sinx = R.H.S

Q21.Prove that

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

Answer.

L.H.S

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

Arranging the term as follows

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

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

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

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

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

= cot3x = R.H.S

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

Applying the following trigonometric identity

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

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

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

L.H.S = R.H.S

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Chapter 7- Co-ordinate geometry	Chapter 15-Probability
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Chapter 1- Chemical reactions and equations	Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt	Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals	Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 15-Environment
Chapter 8- How do organisms reproduce?	Chapter 16-Management of Natural Resources

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	Chapter 16- Probability

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