Q66 / 180 · KCET 2010 · Mathematics
If $a>b>0, \sec ^{-1} \frac{a+b}{a-b}=2 \sin ^{-1} x$, then $x$ is
- A$-\sqrt{\frac{b}{a+b}}$
- Correct answerB$\sqrt{\frac{b}{a+b}}$
- C$-\sqrt{\frac{a}{a+b}}$
- D$\sqrt{\frac{a}{a+b}}$
Solution
If $a>b>0, \sec ^{-1}\left(\frac{a+b}{a-b}\right)=2 \sin ^{-1} x$
$\Rightarrow \quad \cos ^{-1}\left(\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}\right)=2 \sin ^{-1} \mathrm{x}$
$\Rightarrow \quad \cos ^{-1}\left(\frac{1-\frac{b}{a}}{1+\frac{b}{a}}\right)=2 \sin ^{-1} x$
$\Rightarrow \quad \cos ^{-1}\left\{\frac{1-(\sqrt{b / a})^{2}}{1+(\sqrt{b / a})^{2}}\right\}=2 \sin ^{-1} x$
$\Rightarrow \quad 2 \tan ^{-1}(\sqrt{\mathrm{b} / \mathrm{a}})=2 \sin ^{-1} \mathrm{x}$
$\left[\because 2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right]$
$\Rightarrow \quad \sin ^{-1}\left(\frac{\sqrt{b}}{\sqrt{a+b}}\right)=\sin ^{-1} x$
$\left[\because \tan ^{-1} x=\sin ^{-1} \frac{x}{\sqrt{1+x^{2}}}\right]$
$\Rightarrow \quad x=\sqrt{\frac{b}{a+b}}$
Hence, the correct answer is option: 'B'.