Question

If ${\tan }^{-1}\left\{\frac{x}{a+\sqrt{a^2-x^2}}\right\}=\frac{1}{p}{\sin }^{-1}\frac{x}{a}$, $-a<x<a$.
find the value of p.

• A. $p=-1$
• B. $p=+1$
• C. $p=+2$
• D. $p=-2$

Answer 1

Answer: (C)

$p=+2$

Let $x=a\sin \theta$,
Given $-a<x<a$.
Then, $-a<a\sin \theta <a$ or $-1<\sin \theta <1$
$\Rightarrow$ $\theta ϵ(-\frac{\pi }{2},\frac{\pi }{2})$
$\therefore$ ${\tan }^{-1}\left\{\frac{x}{a+\sqrt{a^2-x^2}}\right\}={\tan }^{-1}\left\{\frac{a\sin \theta }{a+\sqrt{a^2-a^2{\sin }^2\theta }}\right\}$
$={\tan }^{-1}\left\{\frac{\sin \theta }{1+\cos \theta }\right\}$

$={\tan }^{-1}\left\{\frac{2\sin \left(\theta /2\right)\cos \left(\theta /2\right)}{2{\cos }^2\left(\theta /2\right)}\right\}$
$={\tan }^{-1}\left\{\tan \frac{\theta }{2}\right\}$
$=\frac{\theta }{2}=\frac{1}{2}{\sin }^{-1}\frac{x}{a}$