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. $\frac{Ï€}{4}$
• B. $\frac{Ï€}{3}$
• C. $\frac{Ï€}{2}$
• D. $Ï€$

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, $T_n=si{n}^{−1}\left(\frac{\sqrt{n}−\sqrt{n−1}}{\sqrt{n\left(n+1\right)}}\right)$
$=si{n}^{−1}\left[\frac{1}{\sqrt{n}}\sqrt{1−{\left(\frac{1}{\sqrt{n+1}}\right)}^2}−\frac{1}{\sqrt{n+1}}\sqrt{1−{\left(\frac{1}{\sqrt{n}}\right)}^2}\right]$
$=si{n}^{−1}\frac{1}{\sqrt{n}}−si{n}^{−1}\frac{1}{\sqrt{n+1}}$
$\left[âˆ\mu si{n}^{−1}x−si{n}^{−1}y=si{n}^{−1}\left(x\sqrt{1−y^2}−y\sqrt{1−x^2}\right)\right]$
$∴S=si{n}^{−1}\frac{1}{\sqrt{2}}+\left(si{n}^{−1}\frac{1}{\sqrt{2}}−sin\frac{1}{\sqrt{3}}\right)$
$+\left(si{n}^{−1}\frac{1}{\sqrt{3}}−si{n}^{−1}\frac{1}{\sqrt{4}}\right)+.....+∞$
$=2si{n}^{−1}\frac{1}{\sqrt{2}}$
$=2\left(\frac{Ï€}{4}\right)$
$=\frac{Ï€}{2}$