Question

The value of ${\sum }_{k=1}^n{\sin }^{-1}\left(\frac{\sqrt{k}-\sqrt{k-1}}{\sqrt{k(k+1)}}\right)$ is

• A. ${\tan }^{-1}\sqrt{n}+\frac{\pi }{4}$
• B. ${\tan }^{-1}\sqrt{n}$
• C. ${\tan }^{-1}\sqrt{n-1}-\frac{\pi }{4}$
• D. None of these

Answer 1

The correct option is B ${\tan }^{-1}\sqrt{n}$

To find $\sum _{k=1}^n{\sin }^{-1}\left(\frac{\sqrt{k}-\sqrt{k-1}}{\sqrt{k(k+1)}}\right)$

Let $t_k={\sin }^{-1}\left(\frac{\sqrt{k}-\sqrt{k-1}}{\sqrt{k(k+1)}}\right)$

$={\sin }^{-1}\left(\frac{1}{\sqrt{k}}\sqrt{\frac{k}{k+1}}-\frac{1}{\sqrt{k+1}}\sqrt{\frac{k-1}{k}}\right)$

$={\sin }^{-1}\left(\frac{1}{\sqrt{k}}\sqrt{1-\frac{1}{k+1}}-\frac{1}{\sqrt{k+1}}\sqrt{1-\frac{1}{k}}\right)$

$t_k={\sin }^{-1}\left(\frac{1}{\sqrt{k}}\right)-{\sin }^{-1}\left(\frac{1}{\sqrt{k+1}}\right)$

Now,

$\sum _{k=1}^n{\sin }^{-1}\left(\frac{\sqrt{k}-\sqrt{k-1}}{\sqrt{k(k+1)}}\right)$

$={\sin }^{-1}\left(1\right)-{\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)+{\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)-{\sin }^{-1}\left(\frac{1}{\sqrt{3}}\right)+$ $.......{\sin }^{-1}\left(\frac{1}{\sqrt{n}}\right)-{\sin }^{-1}\left(\frac{1}{\sqrt{n+1}}\right)$

$={\sin }^{-1}\left(1\right)-{\sin }^{-1}\left(\frac{1}{\sqrt{n+1}}\right)$

$=\frac{\pi }{2}-{\sin }^{-1}\left(\frac{1}{\sqrt{n+1}}\right)$

$={\cos }^{-1}\left(\frac{1}{\sqrt{n+1}}\right)={\tan }^{-1}\sqrt{n}$