∑r=1n sin−1⁡r−r−1r(r+1) is equal to

sin−1⁡r−r−1r(r+1)=tan−1⁡r−r−11+r(r−1)∴∑r=1n sin−1⁡r−r−1r(r+1)=∑r=1n tan−1⁡r − tan−1⁡r−1 =tan−11-tan−10+tan−12-tan−11+.........tan-1n-tan−1n-1=tan-1n

∑r=1n sin−1⁡r−r−1r(r+1) is equal to

sin−1⁡r−r−1r(r+1)=tan−1⁡r−r−11+r(r−1)∴∑r=1n sin−1⁡r−r−1r(r+1)=∑r=1n tan−1⁡r − tan−1⁡r−1 =tan−11-tan−10+tan−12-tan−11+.........tan-1n-tan−1n-1=tan-1n

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$\sum_{r = 1}^{n} \sin^{- 1} (\frac{\sqrt{r} - \sqrt{r - 1}}{\sqrt{r (r + 1)}})$ is equal to

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$\tan^{- 1} (\sqrt{n}) - \frac{π}{4}$

$\tan^{- 1} (\sqrt{n} + 1) - \frac{π}{4}$

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

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

answer is C.

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Detailed Solution

$\begin{array}{l} \sin^{- 1} (\frac{\sqrt{r} - \sqrt{r - 1}}{\sqrt{r} (r + 1)}) = \tan^{- 1} (\frac{\sqrt{r} - \sqrt{r - 1}}{1 + \sqrt{r (r - 1)}}) \\ ∴ \sum_{r = 1}^{n} \sin^{- 1} (\frac{\sqrt{r} - \sqrt{r - 1}}{\sqrt{r (r + 1)}}) = \sum_{r = 1}^{n} (\tan^{- 1} \sqrt{r} - \tan^{- 1} \sqrt{r - 1}) = \tan^{- 1} 1 - \tan^{- 1} 0 + \tan^{- 1} 2 - \tan^{- 1} 1 + . . . . . . . . . \tan^{- 1} \sqrt{n} - \tan^{- 1} \sqrt{n - 1} \\ = \tan^{- 1} \sqrt{n} \end{array}$