Question

$\frac{1}{2}{\cos }^{-1}\frac{[1-x]}{[1+x]}=$

• A. $co{t}^{-1}\sqrt{x}$
• B. ${\tan }^{-1}\sqrt{x}$
• C. ${\tan }^{-1}x$
• D. $co{t}^{-1}x$

Answer 1

The correct option is B

${\tan }^{-1}\sqrt{x}$

Explanation for the correct option:

Let $x={\tan }^2\theta$

$\Rightarrow \theta ={\tan }^{-1}\sqrt{x}$

Now, given expression becomes

$\frac{1}{2}{\cos }^{-1}\frac{[1-x]}{[1+x]}=\frac{1}{2}{\cos }^{-1}\frac{[1–{\tan }^2\theta ]}{[1+{\tan }^2\theta ]}$

$=\frac{1}{2}{\cos }^{-1}\times \cos 2\theta$

$=\frac{2\theta }{2}$

$=\theta$

$=ta{n}^{-1}\sqrt{\mathrm{x}}$

Hence, Option ‘B’ is Correct.