A) \[{{\tan }^{-1}}(\sqrt{n})-\frac{\pi }{4}\]
B) \[{{\tan }^{-1}}(\sqrt{n+1})-\frac{\pi }{4}\]
C) \[{{\tan }^{-1}}(\sqrt{n})\]
D) \[{{\tan }^{-1}}(\sqrt{n+1})\]
Correct Answer: C
Solution :
[c] \[{{\sin }^{-1}}\left( \frac{\sqrt{r}-\sqrt{r-1}}{\sqrt{r(r+1)}} \right)={{\tan }^{-1}}\left( \frac{\sqrt{r}-\sqrt{r-1}}{1+\sqrt{r}\sqrt{(r-1)}} \right)\] \[={{\tan }^{-1}}\sqrt{r}-{{\tan }^{-1}}(\sqrt{r-1})\] \[\Rightarrow \sum\limits_{r=1}^{n}{{{\sin }^{-1}}\left( \frac{\sqrt{r}-\sqrt{r-1}}{\sqrt{r}(r+1)} \right)}\] \[=\sum\limits_{r=1}^{n}{(ta{{n}^{-1}}\sqrt{r}-ta{{n}^{-1}}\sqrt{r-1})=ta{{n}^{-1}}\sqrt{n}}\]
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