∑r=1nsin−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=1nsin−1⁡(r−r−1r(r+1))=∑r=1n(tan−1⁡r−tan−1⁡r−1)=tan−1⁡n

∑r=1nsin−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=1nsin−1⁡(r−r−1r(r+1))=∑r=1n(tan−1⁡r−tan−1⁡r−1)=tan−1⁡n

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

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

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

$\tan^{- 1} (\sqrt{n}) - \frac{π}{4}$

answer is C.

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Detailed Solution

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

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

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

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∑r=1nsin−1⁡(r−r−1r(r+1)) is equal to

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