Inverse Trigonometric Functions Question 149
Question: $ \sum\limits_{r=1}^{n}{{{\sin }^{-1}}}( \frac{\sqrt{r}-\sqrt{r-1}}{\sqrt{r(r+1)}} ) $ is equal to
Options:
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}) $
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Answer:
Correct Answer: C
$ {{\sin }^{-1}}( \frac{\sqrt{r}-\sqrt{r-1}}{\sqrt{r(r+1)}} )={{\tan }^{-1}}( \frac{\sqrt{r}-\sqrt{r-1}}{1+\sqrt{r}\sqrt{(r-1)}} ) $
$ ={{\tan }^{-1}}\sqrt{r}-{{\tan }^{-1}}(\sqrt{r-1}) $
$ \Rightarrow \sum\limits_{r=1}^{n}{{{\sin }^{-1}}( \frac{\sqrt{r}-\sqrt{r-1}}{\sqrt{r}(r+1)} )} $
$ =\sum\limits_{r=1}^{n}{(ta{n^{-1}}\sqrt{r}-ta{n^{-1}}\sqrt{r-1})=ta{n^{-1}}\sqrt{n}} $
BETA
