Question

${\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)+{\sin }^{-1}\left(\frac{\sqrt{2}-1}{\sqrt{6}}\right)+\cdots +{\sin }^{-1}\left(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}}\right)+...\infty =$

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{4}$
• D. None of these

Answer 1

The correct option is B $\frac{\pi }{2}$

${\text{sin}}^{-1}(\frac{1}{\sqrt{2}})+{\text{sin}}^{-1}(\frac{\sqrt{2}-1}{\sqrt{6}})+\cdots +{\text{sin}}^{-1}(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}})=\sum {\text{sin}}^{-1}(\frac{1}{\sqrt{r}}\times \frac{\sqrt{r}}{\sqrt{r+1}}-\frac{\sqrt{r-1}}{\sqrt{r}}\times \frac{1}{\sqrt{r+1}})$

$=\sum ({\sin }^{-1}(\frac{\sqrt{r}}{\sqrt{r+1}})-{\text{sin}}^{-1}(\frac{\sqrt{r-1}}{\sqrt{r}}))$

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

${\text{sin}}^{-1}(\frac{1}{\sqrt{2}})+{\text{sin}}^{-1}(\frac{\sqrt{2}-1}{\sqrt{6}})+\cdots +{\text{sin}}^{-1}(\frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n(n+1)}})+\cdots \infty =\underset{n\to \infty }{lim}{\text{sin}}^{-1}(\frac{1}{\sqrt{1+\frac{1}{n}}})=\frac{\pi }{2}$