Need Solution for R.D.Sharma Maths Class 12 Chapter 3 Inverse Trigonomeric Functions Exercise Mutliple Choice Questions  Question 1 Maths Textbook Solution.

Answer: $\sin 2 \alpha$

Hint: Take tan function to the RHS so as one variable goes free. Try to normalize the denominator, so as under root can be replaced.

Given: $\tan ^{-1}\left\{\frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}\right\}=\alpha, \text { then } x^{2}=?$

Solution:

$\tan ^{-1}\left\{\frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}\right\}=\alpha$

$\tan \alpha=\frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}$

\begin{aligned} &\tan \alpha=\frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}} \times \frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}} \\ &\tan \alpha=\frac{\left(\sqrt{1+x^{2}}\right)^{2}+\left(\sqrt{1-x^{2}}\right)^{2}-2 \sqrt{1+x^{2}} \sqrt{1-x^{2}}}{\left(\sqrt{1+x^{2}}\right)^{2}-\left(\sqrt{1-x^{2}}\right)^{2}} \end{aligned}

$\tan \alpha=\frac{1-\sqrt{1-x^{4}}}{x^{2}}$

$x^{2} \tan \alpha=1-\sqrt{1-x^{4}}$

\begin{aligned} &\sqrt{1-x^{4}}=1-x^{2} \tan \alpha \\ &1-x^{4}=1+x^{4} \tan ^{2} \alpha-2 x^{2} \tan \alpha \\ &x^{4}+x^{4} \tan ^{2} \alpha-2 x^{2} \tan \alpha=0 \\ &x^{2}\left(x^{2} \sec ^{2} \alpha-2 \tan \alpha\right)=0 \\ &x^{2} \sec ^{2} \alpha-2 \tan \alpha=0 \\ &x^{2}=\frac{2 \tan \alpha}{\sec ^{2} \alpha}=2 \sin \alpha \cos \alpha=\sin 2 \alpha \end{aligned}

Mention how $\tan ^{\wedge} 2$ a has been taken as $\\sec ^{\wedge} 2$ a steps has been skipped rectify that include the following steps