Question

Given $0\le x\le \frac{1}{2}$, then the value of $tan\left[{\sin }^{-1}\left(\frac{x}{\surd 2}+\frac{\surd 1-x^2}{\surd 2}\right)-{\sin }^{-1}x\right]$ is

• A. $1$
• B. $\sqrt{3}$
• C. $-1$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is A

$1$

Explanation for the correct option:

Step 1. Find the value of given equation:

For $0\le x\le \frac{1}{2}$,

$tan\left[{\sin }^{-1}\left(\frac{x}{\surd 2}+\frac{\sqrt{1-x^2}}{\surd 2}\right)-{\sin }^{-1}x\right]$

$=tan\left[{\sin }^{-1}\left(\frac{x+\sqrt{1-x^2}}{\surd 2}\right)–{\sin }^{-1}x\right]$

Step 2. Put ${\sin }^{-1}x=\theta$

$\Rightarrow$ $x=\sin \theta$

$=tan\left[{\sin }^{-1}\left(\frac{\sin \theta +\sqrt{[1–{\sin }^2\theta }]}{\surd 2}\right)–\theta \right]$ $\left[\mathbf{\because }\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathbf{+}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\right]$

$=tan\left[{\sin }^{-1}\left(\frac{1}{\surd 2}\sin \theta +\frac{1}{\surd 2}\cos \theta \right)–\theta \right]$

$=tan\left[{\sin }^{-1}\left(\cos \frac{\mathrm{\pi }}{4}\times \sin \theta +\sin \frac{\mathrm{\pi }}{4}\times \cos \theta \right)–\theta \right]$

$=tan\left[{\sin }^{-1}\left\{\sin \left(\theta +\frac{\pi }{4}\right)\right\}–\theta \right]$ $\left[\mathbf{\because }\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{A}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{B}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{A}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{B}\mathbf{}\mathbf{=}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\left(A+B\right)\right]$

$=\tan \left(\theta +\frac{\mathrm{\pi }}{4}–\theta \right)$

$=\tan \frac{\pi }{4}$

$=\mathbf{1}$

Hence, Option ‘A’ is Correct.