Question

If $a>b>0,Se{c}^{-1}\left(\frac{a+b}{a-b}\right)=2Si{n}^{-1}x$ then $x=$ _____

• A. $-\sqrt{\frac{a}{a+b}}$
• B. $\sqrt{\frac{a}{a+b}}$
• C. $-\sqrt{\frac{b}{a+b}}$
• D. $\sqrt{\frac{b}{a+b}}$

Answer 1

Answer: (D)

If $a>b>0,{\sec }^{-1}\left(\frac{a+b}{a-b}\right)=2{\sin }^{-1}x$
$\Rightarrow {\cos }^{-1}\left(\frac{a-b}{a+b}\right)=2{\sin }^{-1}x$
$\Rightarrow {\cos }^{-1}\left(\frac{1-\frac{b}{a}}{1+\frac{b}{a}}\right)=2{\sin }^{-1}x$
$\Rightarrow {\cos }^{-1}\left\{\frac{1-(\sqrt{b/a}{)}^2}{1+(\sqrt{b/a}{)}^2}\right\}=2{\sin }^{-1}x$
$\Rightarrow 2{\tan }^{-1}(\sqrt{b/a})=2{\sin }^{-1}x$
$\left[\because 2{\tan }^{-1}x={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]$
$\Rightarrow {\sin }^{-1}\left(\frac{\sqrt{b}}{\sqrt{a+b}}\right)={\sin }^{-1}x$
$\left[\because {\tan }^{-1}x={\sin }^{-1}\frac{x}{\sqrt{1+x^2}}\right]$
$\Rightarrow x=\sqrt{\frac{b}{a+b}}$