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Please Solve R.D.Sharma Class 12 Chapter 3 Inverse Trigonometric Functions Exercise Multiple Choice Questions Question 23 Maths Textbbok Solution.

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Answer:  \frac{1}{\sqrt{3}}

Hint: \sin 2 y=\left(\frac{2 \tan y}{1+\tan ^{2} y}\right), \cos 2 y=\left(\frac{1-\tan ^{2} y}{1+\tan ^{2} y}\right), \tan 2 y=\left(\frac{2 \tan y}{1-\tan ^{2} y}\right)

Given: 3 \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)-4 \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+2 \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}


\text { Let } \mathrm{x}=\tan y

Then,3 \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)-4 \cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)+2 \tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}

3 \sin ^{-1}(\sin 2 y)-4 \cos ^{-1}(\cos 2 y)+2 \tan ^{-1}(\tan 2 y)=\frac{\pi}{3}

                                                3(2 y)-4(2 y)+2(2 y)=\frac{\pi}{3}

                                               6 y-8 y+4 y=\frac{\pi}{3}

                                                \begin{aligned} &2 y=\frac{\pi}{3} \\ &y=\frac{\pi}{6} \\ &\tan ^{-1} x=\frac{\pi}{6} \end{aligned}

                                                  x=\tan \frac{\pi}{6}


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