Inverse Trigonometric Functions Question 102
Question: The sum of the infinite series $ {{\sin }^{-1}}( \frac{1}{\sqrt{2}} )+{{\sin }^{-1}}( \frac{\sqrt{2}-1}{\sqrt{6}} )+{{\sin }^{-1}}( \frac{\sqrt{3}-\sqrt{2}}{\sqrt{12}} )+… $
$ +…+{{\sin }^{-1}}( \frac{\sqrt{n}-\sqrt{(n-1)}}{\sqrt{{n(n+1)}}} )+… $ is
Options:
A) $ \frac{\pi }{8} $
B) $ \frac{\pi }{4} $
C) $ \frac{\pi }{2} $
D) $ \pi $
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Answer:
Correct Answer: C
$ \because T_{r}={{\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)}} ) $
$ S_{n}=\sum\limits_{r=1}^{n}{{{\tan }^{-1}}( \frac{\sqrt{r}-\sqrt{(r-1)}}{1+\sqrt{r}\sqrt{(r-1)}} )} $
$ =\sum\limits_{r=1}^{n}{{ta{n^{-1}}\sqrt{r}-ta{n^{-1}}\sqrt{(r-1)}}} $
$ ={{\tan }^{-1}}\sqrt{n}-{{\tan }^{-1}}\sqrt{0}={{\tan }^{-1}}\sqrt{n}-0 $
$ \therefore {S_{\infty }}={{\tan }^{-1}}\infty =\frac{\pi }{2} $
BETA
