Question

The sum of the infinite series $\sin^{-1}\frac{1}{\sqrt2}+\sin^{-1}\left(\frac{\sqrt2-1}{\sqrt6}\right)+\sin^{-1}\left(\frac{\sqrt3-\sqrt2}{\sqrt{12}}\right)+...+\sin^{-1}\left(\frac{\sqrt n-\sqrt{n-1}}{\sqrt{n(n+1)}}\right)$ is

• A. $\frac{\pi}{4}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{2}$
• D. $\pi$

Answer 1

The Correct Option is C

$S = sin^{-1} \frac{1}{\sqrt{2}} + sin^{-1} \frac{\sqrt{2}-1}{\sqrt{6}} + sin^{-1} \frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}} + $ $ ..... + sin^{-1} \left(\frac{\sqrt{n} - \sqrt{n-1}}{\sqrt{n\left(n+1\right)}}\right) $ Now, $T_{n}= sin^{-1}\left(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n\left(n+1\right)}}\right)$ $ = sin^{-1} \left[ \frac{1}{\sqrt{n}} \sqrt{1-\left(\frac{1}{\sqrt{n+1}}\right)^{2}} -\frac{1}{\sqrt{n+1}}\sqrt{1-\left(\frac{1}{\sqrt{n}}\right)^{2}}\right]$ $=sin^{-1} \frac{1}{\sqrt{n}} -sin^{-1} \frac{1}{\sqrt{n+1}} $ $ \left[ \because sin^{-1} x -sin^{-1}y = sin^{-1}\left(x\sqrt{1-y^{2}} -y \sqrt{1-x^{2}}\right)\right] $ $ \therefore S = sin^{-1} \frac{1}{\sqrt{2}} + \left(sin ^{-1} \frac{1}{\sqrt{2}} - sin \frac{1}{\sqrt{3}}\right) $ $+ \left(sin^{-1} \frac{1}{\sqrt{3}}-sin^{-1} \frac{1}{\sqrt{4}}\right) + .....+\infty $ $ = 2sin^{-1} \frac{1}{\sqrt{2}} $ $= 2\left(\frac{\pi}{4}\right) $ $= \frac{\pi}{2}$