Question

$\sum _{r=1}^n{\sin }^{-1}\left(\frac{\sqrt{r}-\sqrt{r-1}}{\sqrt{r(r+1)}}\right)$ is equal to

• A. ${\tan }^{-1}\left(\sqrt{n}\right)-\frac{\pi }{4}$
• B. ${\tan }^{-1}\left(\sqrt{n+1}\right)-\frac{\pi }{4}$
• C. ${\tan }^{-1}\left(\sqrt{n}\right)$
• D. ${\tan }^{-1}\left(\sqrt{n+1}\right)$

Answer 1

Answer: (C)

${\tan }^{-1}\left(\sqrt{n}\right)$

$\sum _{r=1}^n{\sin }^{-1}\left(\frac{\sqrt{r}-\sqrt{r-1}}{\sqrt{r(r+1)}}\right)=\sum _{r=1}^n{\tan }^{-1}\left(\frac{\frac{\sqrt{r}-\sqrt{r-1}}{\sqrt{r(r+1)}}}{\sqrt{1-{\left(\frac{\sqrt{r}-\sqrt{r-1}}{\sqrt{r(r+1)}}\right)}^2}}\right)$

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

$=({\tan }^{-1}\sqrt{1}-{\tan }^{-1}\sqrt{0})+({\tan }^{-1}\sqrt{2}-{\tan }^{-1}\sqrt{1})$

$+({\tan }^{-1}\sqrt{3}-{\tan }^{-1}\sqrt{2})+...+({\tan }^{-1}\sqrt{n}-{\tan }^{-1}\sqrt{n-1})$

$={\tan }^{-1}\sqrt{n}-0={\tan }^{-1}\sqrt{n}$