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Class 12 NCERT Solutions- Mathematics Part I – Chapter 2 Inverse Trigonometric Functions – Miscellaneous Exercise on Chapter 2

Last Updated : 03 Apr, 2024
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Question 1. Find the value of [Tex]{\cos }^{-1}(\cos \frac {13\pi} {6})[/Tex]


We know that [Tex]\cos^{-1} (\cos x)=x [/Tex] 

Here, [Tex]\frac {13\pi} {6} \notin [0,\pi].[/Tex]

Now, [Tex]{\cos }^{-1}(\cos \frac {13\pi} {6}) [/Tex] can be written as :

[Tex]{\cos }^{-1}(\cos \frac {13\pi} {6})={\cos }^{-1}[\cos( 2\pi+\frac {\pi} {6})] [/Tex], where [Tex]\frac{\pi} {6} \in [0,\pi].[/Tex]

Hence, the value of [Tex]{\cos }^{-1}(\cos \frac {13\pi} {6}) [/Tex] = π/6

Question 2. Find the value of [Tex]\tan^{-1}(\tan \frac {7\pi}{6})[/Tex]


We know that [Tex]\tan^{-1} (\tan x)=x[/Tex]

Here, [Tex]\frac {7\pi}{6} \notin(\frac{-\pi}{2},\frac{\pi}{2})[/Tex]

Now, [Tex]\tan^{-1}(\tan \frac {7\pi}{6}) [/Tex] can be written as:

[Tex]\tan^{-1}(\tan \frac {7\pi}{6})=\tan^{-1}[\tan( 2\pi -\frac {5\pi}{6})]         [/Tex]      [Tex]-[\tan(2\pi-x)=-\tan x][/Tex]

[Tex]\tan^{-1}[-\tan(\frac {5\pi}{6}) ]=\tan^{-1}[\tan(-\frac {5\pi}{6})][/Tex]

[Tex]=\tan^{-1}[\tan(\pi-\frac {5\pi}{6})]=\tan^{-1}[\tan(  \frac {\pi}{6})], [/Tex] where [Tex]\frac{\pi}{6} \in (\frac{-\pi}{2},\frac{\pi}{2})[/Tex]

Hence, the value of [Tex]\tan^{-1}(\tan\frac{7\pi}{6}) [/Tex] = π/6

Question 3. Prove [Tex]2\sin^{-1} \frac{3}{5}=\tan^{-1} \frac{24}{7}[/Tex]


Let [Tex]\sin^{-1} \frac{3}{5}=x [/Tex]          -(1)

sin x = 3/5

So,[Tex] \cos x =  \sqrt{1-(\frac{3}{5})^2 }  [/Tex]= 4/5

tan x = 3/4

Hence, [Tex]x=\tan^{-1}  \frac{3}{4} [/Tex]

Now put the value of x from eq(1), we get

[Tex]\sin^{-1} \frac{3}{5}=\tan^{-1} \frac{3}{4}[/Tex]

Now, we have

L.H.S [Tex]= 2 \sin^{-1} \frac{3}{5}=2 \tan^{-1} \frac{3}{4}[/Tex]

= [Tex]\tan^{-1}(\frac{2 \times \frac{3}{4}}{1-(\frac{3}{4})^{2}}) [/Tex]                -[Tex][2\tan^{-1} x=\tan^{-1} \frac{2x}{1-x^2}][/Tex]

[Tex]= \tan^{-1}(\frac{ \frac{3}{2}}{\frac{16-9}{16}})=\tan^{-1} (\frac{3}{2} \times \frac{16}{7})[/Tex]

[Tex]=\tan^{-1} \frac{24}{7} [/Tex]

Hence, proved.

Question 4. Prove [Tex]\sin^{-1} \frac{8}{17}+\sin^{-1} \frac{3}{5}=\tan^{-1} \frac{77}{36}[/Tex]


Let [Tex]\sin^{-1} \frac{8}{17}=x [/Tex] 

Then sin x = 8/17 

 cos x =[Tex] \sqrt{1-(\frac{8}{17})^2}=\sqrt \frac{225}{289} [/Tex] = 15/17

Therefore, [Tex]\tan x=\frac{8}{15}\implies x=\tan^{-1}\frac{8}{15}[/Tex]

[Tex]\sin^{-1} \frac{8}{17}=\tan^{-1} \frac{8}{17}       [/Tex]         -(1)

Now, let [Tex]\sin^{-1} \frac{3}{5}=y   [/Tex]

Then, sin y = 3/5 

[Tex]\cos y=\sqrt{1- (\frac{3}{5})^2}=\sqrt{ (\frac{16}{25})} [/Tex] = 4/5

[Tex]\therefore \tan y =\frac{3}{4} \implies y=\tan^{-1} \frac{3}{4}[/Tex]

[Tex]\therefore \sin^{-1} \frac{3}{5}=\tan^{-1} \frac{3}{4}      [/Tex]     -(2)

Now, we have:

L.H.S.[Tex]=\sin^{-1} \frac{8}{17}+\sin^{-1} \frac{3}{5}[/Tex]

 From equation(1) and (2), we get

= [Tex]\tan^{-1} \frac{8}{15}+\tan^{-1} \frac{3}{4}  [/Tex]                         

= [Tex]\tan^{-1} \frac{{\frac{8}{15}+ \frac{3}{4}} }{1-{\frac{8}{15}\times \frac{3}{4}}}[/Tex]

= [Tex]\tan^{-1}(\frac{32+45}{60-24}) [/Tex]                                 -[Tex][\tan^{-1} x + \tan^{-1} y=\tan^{-1} \frac{x+y}{1-xy}] [/Tex]

= [Tex]\tan^{-1} \frac{77}{36}    [/Tex]

Hence proved

Question 5. Prove [Tex]\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65}[/Tex]


Let [Tex]\cos^{-1}\frac{4}{5}= x     [/Tex]

Then, cos x = 4/5

[Tex]\sin x = \sqrt {1- (\frac{4}{5})^{2}} [/Tex] = 3/5

[Tex]\therefore \tan x =\frac{3}{4} \implies x=\tan^{-1} \frac{3}{4}[/Tex]

[Tex]\therefore \cos^{-1} \frac{4}{5}=\tan^{-1} \frac{3}{4}      [/Tex]        -(1)

Now let [Tex]\cos^{-1} \frac{12}{13}=x   [/Tex]

Then, cos y = 3/4 

[Tex]\sin^{-1} y=\frac{5}{13}[/Tex]

[Tex]\therefore\tan y= \frac{5}{12} \implies y=\tan^{-1} \frac{5}{12}[/Tex]

[Tex]\therefore \cos ^{-1} \frac{12}{13}=\tan^{-1} \frac{5}{12}      [/Tex]            -(2)

Let [Tex]\cos^{-1} \frac{33}{65}=z [/Tex] 

Then, cos z = 33/65

sin z = 56/65 

[Tex]\therefore \tan z = \frac{56}{65} \implies z= tan^{-1}\frac{56}{33}[/Tex]

[Tex]\therefore \cos^{-1} \frac{33}{65}= \tan^{-1} \frac{56}{33}      [/Tex]    -(3)

Now, we will prove that : 

L.H.S. [Tex]=\cos^{-1} \frac{3}{5}\cos^{-1} \frac{12}{13}[/Tex]

From equation (1) and equation (2)

= [Tex]\tan^{-1} \frac{3}{4}+\tan^{-1} \frac{5}{12}      [/Tex]                                                                 

= [Tex]\tan^{-1} \frac{{\frac{3}{4}+ \frac{5}{12}} }{1-{\frac{3}{4}\times \frac{5}{12}}}      [/Tex]         -[Tex][\tan^{-1} x +\tan^{-1} y=\tan^{-1} \frac{x+y}{1-xy}][/Tex]

= [Tex]\tan^{-1} \frac{36+20}{48-15}[/Tex]

= [Tex]\tan^{-1} \frac{56}{33}[/Tex]

Using equation(3)

= [Tex]\tan^{-1} \frac{56}{33}      [/Tex]                                                                                                  

Hence proved

Question 6. Prove [Tex]\cos^{-1} \frac{12}{13}+\sin^{-1} \frac{3}{5}=\sin^{-1} \frac{56}{65}[/Tex]


Let [Tex]\sin^{-1} \frac{3}{5}=x [/Tex]  

Then, sin x = 3/5 

[Tex]\cos x =\sqrt{1- (\frac{3}{5})^{2}}=\sqrt \frac{16}{25} [/Tex] = 4/5

[Tex]\therefore \tan x = \frac{3}{4} \implies x= \tan^{-1} \frac{3}{4}[/Tex]

[Tex]\therefore \sin^{-1} \frac{3}{5}= \tan^{-1} \frac{3}{4}     [/Tex]           -(1)

Now, let [Tex]\cos^{-1} \frac{12}{13}=y [/Tex]  

Then, cos y = 12/13 and sin y = 5/13

[Tex]\therefore \tan y = \frac{5}{12} \implies y= \tan^{-1} \frac{5}{12}[/Tex]

[Tex]\therefore \cos^{-1} \frac{12}{13}= \tan^{-1} \frac{5}{12}     [/Tex]        -(2)                                                             

Let [Tex]\sin^{-1} \frac{56}{65}=z[/Tex]

Then, sin z = 56/65 and cos z = 33/65

[Tex]\therefore \tan z = \frac{56}{33} \implies z=\tan ^{-1} \frac{56}{33}[/Tex]

[Tex]\therefore \sin^{-1} \frac{56}{65}= \tan^{-1} \frac{56}{33}     [/Tex]           -(3)

Now, we have:

L.H.S.=[Tex]\cos^{-1} \frac{12}{13}+ \sin^{-1} \frac{3}{5}[/Tex]

From equation(1) and equation(2)

=[Tex]\tan^{-1} \frac{5}{12}+\tan^{-1} \frac{3}{4}     [/Tex]                                                                           

= [Tex]\tan^{-1} \frac{{\frac{5}{12}+ \frac{3}{4}} }{1-{\frac{5}{12}\times \frac{3}{4}}}  [/Tex]               -[Tex][\tan^{-1} x +\tan^{-1} y=\tan^{-1} \frac{x+y}{1-xy}] [/Tex]

= [Tex]\tan^{-1} \frac{20+36}{48-15}[/Tex]

= [Tex]\tan^{-1} \frac{56}{33}[/Tex]

From equation (3)

= [Tex]\sin^{-1} \frac{56}{65} [/Tex]    

Hence proved                                                                                                           

Question 7. Prove [Tex]\tan^{-1} \frac{63}{16}= \sin^{-1} \frac{5}{13}+\cos^{-1} \frac{3}{5}[/Tex]


Let [Tex]\sin^{-1} \frac{5}{13}=x [/Tex]   

Then, sin x = 5/13 and cos x = 12/13.

[Tex]\tan^{-1} \frac{7+5}{35-1}+\tan^{-1} \frac{8+3}{24-1}[/Tex]

[Tex]\therefore \tan x= \frac{5}{12} \to x= \tan^{-1} \frac{5}{12}[/Tex]

[Tex]\therefore \sin^{-1} \frac{5}{13}= \tan^{-1} \frac{5}{12}     [/Tex]                   -(1)

Let [Tex]\cos^{-1} \frac{3}{5}=y [/Tex]     

Then, cos y = 3/5 and sin y = 4/5

[Tex]\therefore \tan y= \frac{4}{3} \implies  y= \tan^{-1}\frac{4}{3}[/Tex]

[Tex]\therefore \cos ^{-1}\frac{3}{5}=\tan^{-1} \frac{4}{3}     [/Tex]                -(2)

From equation(1) and (2), we have

R.H.S.[Tex]=\sin^{-1} \frac{5}{13}+\cos^{-1} \frac{3}{5}[/Tex]

=[Tex]\tan^{-1} \frac{5}{12}+\tan^{-1} \frac{4}{3}[/Tex]

= [Tex]\tan^{-1} \frac{{\frac{5}{12}+ \frac{4}{3}} }{1-{\frac{5}{12}\times \frac{4}{3}}}      [/Tex]          -[Tex][\tan^{-1} x +\tan^{-1} y=\tan^{-1} \frac{x+y}{1-xy}][/Tex]

=[Tex]\tan^{-1} \frac{15+48}{36-20}[/Tex]

=[Tex]\tan^{-1} \frac{63}{16}[/Tex]

L.H.S = R.H.S

Hence proved

Question 9. Prove [Tex]\tan^{-1} \sqrt x= \frac{1}{2} \cos^{-1} (\frac{1-x}{1+x}),x\in[0,1][/Tex]


Let x = tan2θ

Then,[Tex]\sqrt x=\tan \theta \implies\theta=\tan^{-1} \sqrt x.[/Tex]

[Tex]\therefore  \frac{1-x}{1+x}+\frac{1-\tan^{2}\theta}{1+\tan^{2}\theta}=\cos 2\theta[/Tex]

Now, we have

R.H.S = [Tex]\frac{1}{2} \cos ^{-1}(\frac{1-x}{1+x})= \frac{1}{2} \cos ^{-1} (\cos 2 \theta)=\frac{1}{2} \times 2 \theta=\theta=\tan^{-1}\sqrt x[/Tex]

L.H.S = R.H.S

Hence proved

Question 9. Prove [Tex]\cot^{-1} (\frac{\sqrt ({1+ \sin x}) + \sqrt ({1-\sin x})} {\sqrt ({1+ \sin x})-\sqrt ({1- \sin x})})=\frac{x}{2},x\in(0,\frac{\pi}{4})[/Tex]


Consider[Tex] (\frac{\sqrt ({1+ \sin x}) + \sqrt ({1-\sin x})} {\sqrt ({1+ \sin x})-\sqrt ({1- \sin x})})[/Tex]

By rationalizing

=[Tex] \frac{(\sqrt ({1+ \sin x}) + \sqrt ({1-\sin x}))^{2}}{(\sqrt{ ({1+ \sin x})}-\sqrt({1- \sin x}))^{2}}     [/Tex]                         

=[Tex] \frac{( {1+ \sin x)} +  {(1-\sin x)} + 2 \sqrt{(1+\sin x)(1-\sin x)}}{{ {1+ \sin x}}-{1+\sin x}}[/Tex]

=[Tex]\frac{2(1+\sqrt{1-\sin^{2}})}{2\sin x}=\frac{1+\cos x}{\sin x}=\frac{2\cos  ^{2}\frac{x}{2}}{2\sin \frac{x}{2}\cos\frac{x}{2}} [/Tex]

= [Tex]\cot \frac{x}{2}[/Tex]

L.H.S = [Tex]\cot^{-1} (\frac{\sqrt ({1+ \sin x}) + \sqrt ({1-\sin x})} {\sqrt ({1+ \sin x})-\sqrt ({1- \sin x})})={\cot}^{-1}({\cot( \frac x 2)})  [/Tex]= x/2

L.H.S = R.H.S

Hence proved

Question 10. Prove [Tex]\tan^{-1}(\frac{\sqrt{(1+ x )}-\sqrt{(1-x)}}{\sqrt{(1+ x) }+\sqrt{(1-x)}})=\frac{\pi}{4}-\frac{1}{2} \cos^{-1}x,-\frac{1}{\sqrt2}\le x \le1[/Tex]


Put [Tex]x=\cos2\theta [/Tex] so that, [Tex]\theta= \frac{1}{2} \cos^{-1} x.[/Tex]

Then, we have :

LHS = [Tex]\tan^{-1}(\frac{\sqrt{(1+ x) }-\sqrt{(1-x)}}{\sqrt{(1+ x )}+\sqrt{(1-x)}})[/Tex]

= [Tex]\tan^{-1}(\frac{\sqrt{(1+ \cos2 \theta) }-\sqrt{(1-\cos2 \theta)}}{\sqrt{(1+ \cos2 \theta) }+\sqrt{(1-\cos2 \theta)}})[/Tex]

= [Tex]\tan^{-1}(\frac{\sqrt{(2 \cos^{2} \theta) }-\sqrt{(2 \sin^{2}\theta)}}{\sqrt{(2 \cos^{2}\theta) }+\sqrt{(2 \sin^{2} \theta)}})[/Tex]

= [Tex]\tan^{-1}(\frac{\sqrt{2 }\cos \theta -{\sqrt{2 }\cos\theta}}{{\sqrt{2 }\cos\theta }+{\sqrt{2 }\cos \theta}})[/Tex]

= [Tex]\tan^{-1}(\frac{\cos \theta – \sin \theta}{\cos \theta + \sin \theta})=\tan^{-1}(\frac{1 – \tan \theta}{1 + \tan \theta})[/Tex]

[Tex]\tan^{-1} 1- \tan^{-1}(\tan \theta)      [/Tex]      – [Tex][\tan^{-1}(\frac{x-y}{1+xy})]=\tan^{-1}x+\tan^{-1}y[/Tex]

[Tex]=\frac{\pi}{4} -\theta = \frac{\pi}{4}-\frac{1}{2} \cos^{-1} x      [/Tex]  

L.H.S = R.H.S

Hence Proved

Question 11. Solve [Tex]2\tan^{-1}(\cos x)=\tan^{-1}(2\cosec x)[/Tex]


[Tex]2\tan^{-1}(\cos x)=\tan^{-1}(2\cosec x)[/Tex]

=[Tex] \tan^{-1}(\frac{2 \cos x}{1- \cos^{2}x})=\tan^{-1} (2\cosec x)   [/Tex]                      -[Tex][2 \tan^{-1} x=\tan^{-1} \frac{2x}{1-x^{2}}] [/Tex]

= [Tex]\frac{2\cos x} {1-cos^{2}x}=2\cosec x[/Tex]

= [Tex]\frac{ 2 \cos x}{\sin^{2}}= \frac{2}{\sin x}[/Tex]

= cos x/sin x

= cot x =1

Therefore, x = π/4

Question 12. Solve [Tex]\tan^{-1} \frac{1-x}{1+x}=\frac{1}{2} \tan^{-1} x,(x>0)[/Tex]


[Tex]\tan^{-1} \frac{1-x}{1+x}=\frac{1}{2} \tan^{-1} x[/Tex]

Let x = tanθ

[Tex]\tan^{-1}(\frac{1-tan\theta}{1+tan\theta})=\frac{1}{2} \tan^{-1} tan\theta     [/Tex]        

[Tex]\tan^{-1}(\frac{tan\frac{\pi}{4}-tan\theta}{tan\frac{\pi}{4}+tan\theta})=\frac{1}{2} \theta    [/Tex]

[Tex]\tan^{-1}tan(\frac{\pi}{4}-\theta)=\frac{\theta }{2}    [/Tex]

π/4 – θ = θ/2

θ = π/6

So, x = tan(π/6) = 1/√3

Question 13. Solve [Tex]\sin(\tan^{-1}x),|x|<1 [/Tex] is equal to

(A) [Tex]\frac{x}{\sqrt{(1-x^{2})}} [/Tex]    (B) [Tex]\frac{1}{\sqrt{(1-x^{2})}}     [/Tex] (C) [Tex]\frac{1}{\sqrt{(1+x^{2})}}     [/Tex](D) [Tex]\frac{x}{\sqrt{(1+x^{2})}}[/Tex]


Let tan y = x, [Tex]\sin y = \frac{x}{\sqrt{(1+x^{2})}}[/Tex]

Let [Tex]\tan^{-1} x=y [/Tex] Then,

[Tex]\therefore y=\sin^{-1}(\frac{x}{\sqrt{(1+x^{2})}}) \implies \tan^{-1}x=\sin^{-1}\frac{x}{\sqrt{(1+x^{2})}}[/Tex]

[Tex]\therefore \sin(\tan^{-1}x)=sin(sin^{-1}\frac{x}{\sqrt{(1+x^{2})}})=\frac{x}{\sqrt{(1+x^{2})}}[/Tex]

So, the correct answer is D.

Question 14. Solve [Tex]\sin^{-1}(1- x)-2 \sin^{-1}x=\frac{\pi}{2} [/Tex], then x is equal to

(A) 0, 1/2      (B) 1, 1/2      (C) 0     (D) 1/2 


[Tex]\sin^{-1}(1- x)-2 \sin^{-1}x=\frac{\pi}{2}[/Tex]

[Tex]\implies -2 \sin^{-1}x=\frac{\pi}{2} – \sin^{-1}(1-x)[/Tex]

[Tex]\implies -2 \sin^{-1}x=\cos^{-1}(1-x)     [/Tex]        -(1)

Let [Tex]\sin^{-1} x =\theta \to \sin \theta=x [/Tex]

[Tex]\cos \theta= \sqrt{1-x^{2}}[/Tex]

[Tex]\therefore \theta= \cos^{-1}(\sqrt{1-x^{2}})[/Tex]

[Tex]\therefore \sin^{-1} x=cos^{-1}(\sqrt{1-x^{2}})[/Tex]

Therefore, from equation(1), we have


Put x = siny then, we have:

[Tex]-2\cos^{-1}(\sqrt{1-\sin^{2} y})=\cos^{-1}(1-\sin y)[/Tex]

[Tex]-2 \cos^{-1}(\cos y)=\cos^{-1}(1-\sin y)[/Tex]

[Tex]-2 y=\cos^{-1}(1-\sin y)[/Tex]

[Tex]1- \sin y=\cos (-2y)=\cos 2y[/Tex]

[Tex]1- \sin y= 1- 2 \sin^{2} y[/Tex]

[Tex]2\sin^{2} y- \sin y=0[/Tex]

[Tex]\sin y(2 \sin y-1)=0[/Tex]

sin y = 0 or 1/2

x = 0 or x = 1/2

But, when x = 1/2 it can be observed that:

L.H.S. = [Tex]\sin^{-1}(1-\frac{1}{2}) -2\sin^{-1} \frac{1}{2}[/Tex]

= [Tex]\sin^{-1} (\frac{1}{2})-2\sin^{-1} \frac{1}{2}[/Tex]

= [Tex]-\sin^{-1} \frac{1}{2}[/Tex]

= [Tex]- \frac{\pi}{6} \ne\frac{\pi}{2}\ne \space R.H.S.[/Tex]

x = 1/2 is not the solution of given equation.

Thus, x = 0 

Hence, the correct answer is C

Prove [Tex]\tan^{-1} \frac{1}{5}+\tan^{-1} \frac{1}{7}\tan^{-1} \frac{1}{3}+\tan^{-1} \frac{1}{8}=\frac{\pi}{4}[/Tex]


L.H.S.[Tex]=\tan^{-1} \frac{1}{5}+\tan^{-1} \frac{1}{7}\tan^{-1} \frac{1}{3}+\tan^{-1} \frac{1}{8}[/Tex]

= [Tex]\tan^{-1} \frac{{\frac{1}{5}+ \frac{1}{7}} }{1-{\frac{1}{5}\times \frac{1}{7}}} +\tan^{-1} \frac{{\frac{1}{3}+ \frac{1}{8}} }{1-{\frac{1}{3}\times \frac{1}{8}}}  [/Tex]     -[Tex][\tan^{-1} x +\tan^{-1} y=\tan^{-1} \frac{x+y}{1-xy}][/Tex]

= [Tex]\tan^{-1} \frac{7+5}{35-1}+\tan^{-1} \frac{8+3}{24-1}[/Tex]

= [Tex]\tan^{-1} \frac{12}{34}+\tan^{-1} \frac{11}{23}[/Tex]

= [Tex]\tan^{-1} \frac{6}{17}+\tan^{-1} \frac{11}{23}[/Tex]

= [Tex]\tan^{-1} \frac{{\frac{6}{17}+ \frac{11}{23}} }{1-{\frac{6}{17}\times \frac{11}{23}}} [/Tex]

= [Tex]\tan^{-1} \frac{138 + 187}{391-66}[/Tex]

= [Tex]\tan^{-1} \frac{325}{325}=\tan^{-1} 1[/Tex]

= π/4 

L.H.S = R.H.S

Hence proved

Prove [Tex]\frac{9\pi}{8}- \frac{9}{4} \sin^{-1} \frac{1}{3}=\frac{9}{4} \sin^{-1} \frac{2\sqrt2}{3}[/Tex]


L.H.S. = [Tex]\frac{9\pi}{8}- \frac{9}{4} \sin^{-1} \frac{1}{3}[/Tex]

= [Tex]\frac {9}{4}(\frac{\pi}{2}- \sin^{-1} \frac{1}{3})[/Tex]

Using [Tex]\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}[/Tex]

= [Tex]\frac {9}{4}(\cos^{-1} \frac{1}{3})      [/Tex]      -(1)

Now, let [Tex]\cos^{-1}\frac{1}{3}=x [/Tex] Then, [Tex]\cos x =\frac{1}{3} \implies \sin x=\sqrt{1-(\frac{1}{3})^{2}}=\frac{2\sqrt{2}}{3}   [/Tex]

[Tex]\therefore x=\sin^{-1} \frac{2\sqrt2}{3} [/Tex] 

Using equation(1), we get,

= [Tex]\frac{9}{4}\sin^{-1} \frac{2\sqrt2}{3}[/Tex]

L.H.S = R.H.S

Hence Proved

Solve [Tex]\tan ^{-1}(\frac{x}{y})-\tan ^{-1}(\frac{x-y}{x+y}) [/Tex] is equal to

(A) π​/2      (B) π​/3     (C) π​/4      (D) -3π​/4  


[Tex]\tan^{-1} (\frac{x}{y})-\tan^{-1} \frac{x-y}{x+y}[/Tex]

[Tex]\tan^{-1}[\frac{\frac{x}{y}-\frac{x-y}{x+y}}{1+\frac{x}{y} \times\frac{x-y}{x+y}}]     [/Tex]                -[Tex][\tan^{-1}x+\tan^{-1}y=[\tan^{-1}(\frac{x-y}{1+xy})]][/Tex]

[Tex]\tan^{-1}[\frac {\frac {x(x+y)-y(x-y)} {y(x+y)} } {\frac {y(x+y)+x(x-y)} {y(x+y)}}]   [/Tex]

[Tex]{\tan}^{-1}[{\frac {x^2+xy-xy+y^2} {xy+y^2+x^2-xy}}][/Tex]

[Tex]{\tan}^{-1}[\frac {x^2+y^2} {x^2+y^2}]=tan^{-1}1=\frac {\pi} {4}[/Tex]

Hence, the correct answer is C

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