Question

Differential coefficient of $co{t}^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ with respect to ${\tan }^{-1}\left(\frac{2x\sqrt{1-x^2}}{2x^2-1}\right)$, where $x\in [0,\frac{1}{\sqrt{2}})$, is

• A. $\frac{1-x^2}{1+x^2}$
• B. $\frac{1+x^2}{1-x^2}$
• C. $\frac{1}{4}.\frac{\sqrt{1-x^2}}{1+x^2}$
• D. none of these

Answer 1

The correct option is C $\frac{1}{4}.\frac{\sqrt{1-x^2}}{1+x^2}$

For the first one let $x=tanA$
Therefore upon simplification we get that
$co{t}^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$
$=co{t}^{-1}tan\left(A/2\right)$
$=\frac{\pi }{2}-\frac{A}{2}$
$=\frac{\pi -ta{n}^{-1}x}{2}$
$=u$ ...(i)
Substituting $x=cosB$ in the second one and simplifying we get
$ta{n}^{-1}\left(tan2B\right)$
$=2A$
$=2co{s}^{1}x$
$=v$
Hence
$\frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}=\frac{\frac{-1}{1+x^2}}{\frac{-2}{\sqrt{1-x^2}}}=\frac{\sqrt{1-x^2}}{4(1+x^2)}$