Question

Evaluate ${\tan }^{-1}\left[\frac{\sqrt{1+{\mathrm{x}}^2}+\sqrt{1–{\mathrm{x}}^2}}{\sqrt{1+{\mathrm{x}}^2}-\sqrt{1–{\mathrm{x}}^2}}\right]$

• A. $\frac{\mathrm{\pi }}{4}+\frac{1}{2}{\cos }^{-1}{\mathrm{x}}^2$
• B. $\frac{\mathrm{\pi }}{4}+{\cos }^{-1}{\mathrm{x}}^2$
• C. $\frac{\mathrm{\pi }}{4}+\frac{1}{2}{\cos }^{-1}\mathrm{x}$
• D. $\frac{\mathrm{\pi }}{4}-\frac{1}{2}{\cos }^{-1}{\mathrm{x}}^2$

Answer 1

The correct option is A

$\frac{\mathrm{\pi }}{4}+\frac{1}{2}{\cos }^{-1}{\mathrm{x}}^2$

Explanation for correct option

Simplifying the given expression:

$$\begin{gathered} {\tan }^{-1}\left[\frac{\sqrt{1+{\mathrm{x}}^2}+\sqrt{1–{\mathrm{x}}^2}}{\sqrt{1+{\mathrm{x}}^2}-\sqrt{1–{\mathrm{x}}^2}}\right] \\ ={\tan }^{-1}\left[\frac{\sqrt{1+\cos 2\mathrm{\alpha }}+\sqrt{1-\cos 2\mathrm{\alpha }}}{\sqrt{1+\cos 2\mathrm{\alpha }}-\sqrt{1-\cos 2\mathrm{\alpha }}}\right]\left[\mathrm{Let}{x}^2=\cos 2\alpha \right] \end{gathered}$$

We know that:

$$\begin{gathered} \cos 2\theta =2{\cos }^2\theta -1 \\ \Rightarrow \mathbf{}\mathbf{1}\mathbf{+}\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{\theta }\mathbf{=}\mathbf{2}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{\theta } \\ \therefore \sqrt{1+\cos 2\mathrm{\alpha }}=\sqrt{2{\cos }^2\mathrm{\alpha }}=\sqrt{2}\mathrm{cos\alpha }------\left(1\right) \end{gathered}$$

Also,

$$\begin{gathered} \cos 2\mathrm{\theta }=1-2{\sin }^2\mathrm{\theta } \\ \Rightarrow \mathbf{1}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathbf{2}\mathit{\theta }\mathbf{=}\mathbf{2}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{\theta } \\ \therefore \sqrt{1-\cos 2\mathrm{\alpha }}=\sqrt{2{\sin }^2\mathrm{\alpha }}=\sqrt{2}\mathrm{sin\alpha }-----\left(2\right) \end{gathered}$$

So, we get:

$$\begin{gathered} {\tan }^{-1}\left[\frac{\sqrt{2}\mathrm{cos\alpha }+\sqrt{2}\mathrm{sin\alpha }}{\sqrt{2}\mathrm{cos\alpha }-\sqrt{2}\mathrm{sin\alpha }}\right]={\tan }^{-1}\left(\frac{\frac{\sqrt{2}\cos \alpha }{\sqrt{2}\mathrm{cos\alpha }}+\frac{\sqrt{2}\mathrm{sin\alpha }}{\sqrt{2}\mathrm{cos\alpha }}}{{\displaystyle \frac{\sqrt{2}\mathrm{cos\alpha }}{\sqrt{2}\mathrm{cos\alpha }}}-{\displaystyle \frac{\sqrt{2}\mathrm{sin\alpha }}{\sqrt{2}\mathrm{cos\alpha }}}}\right) \\ ={\tan }^{-1}\left(\frac{1+\mathrm{tan\alpha }}{1-\mathrm{tan\alpha }}\right) \\ ={\tan }^{-1}\left(\frac{\tan \frac{\mathrm{\pi }}{4}+\mathrm{tan\alpha }}{1-\tan \frac{\mathrm{\pi }}{4}\mathrm{tan\alpha }}\right)\left[\because \tan \frac{\pi }{4}=1\right] \\ ={\tan }^{-1}\left(\tan \left(\frac{\mathrm{\pi }}{4}+\mathrm{\alpha }\right)\right) \\ =\frac{\mathrm{\pi }}{4}+\mathrm{\alpha } \\ =\frac{\mathrm{\pi }}{4}+\frac{1}{2}{\cos }^{-1}{\mathrm{x}}^2 \end{gathered}$$

Hence, the correct answer is option (A).