Question

The sum of the infinite series ${\sin }^{-1}\frac{1}{\sqrt{2}}+{\sin }^{-1}\left(\frac{\sqrt{2}-1}{\sqrt{6}}\right)+{\sin }^{-1}\left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}}\right)+...+{\sin }^{-1}\left(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}}\right)$ is

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

Answer 1

Answer: (C)

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