∑r=1n sin-1 r-r-1r(r+1) is equal to

∑r=1n sin−1⁡r−r−1r(r+1)=∑r=1n tan−1⁡r−r−11+r⋅r−1=∑r=1n tan−1⁡(r)−tan−1⁡(r−1)=tan−1⁡(1)−tan−1⁡(0)+tan−1⁡(2)−tan−1⁡(1)+tan−1⁡(3)−tan−1⁡(2)+tan−1⁡(4)−tan−1⁡(3)+…+tan−1⁡(n)−tan−1⁡(n−1)=tan−1⁡(n)

∑r=1n sin-1 r-r-1r(r+1) is equal to

∑r=1n sin−1⁡r−r−1r(r+1)=∑r=1n tan−1⁡r−r−11+r⋅r−1=∑r=1n tan−1⁡(r)−tan−1⁡(r−1)=tan−1⁡(1)−tan−1⁡(0)+tan−1⁡(2)−tan−1⁡(1)+tan−1⁡(3)−tan−1⁡(2)+tan−1⁡(4)−tan−1⁡(3)+…+tan−1⁡(n)−tan−1⁡(n−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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$\tan^{- 1} (\sqrt{n + 1}) - \frac{π}{4}$

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

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

$\tan^{- 1} (\sqrt{n}) - \frac{π}{4}$

answer is C.

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Detailed Solution

$\begin{array}{l} \sum_{r = 1}^{n} \sin^{- 1} (\frac{\sqrt{r} - \sqrt{r - 1}}{\sqrt{r (r + 1)}}) \\ = \sum_{r = 1}^{n} \tan^{- 1} (\frac{\sqrt{r} - \sqrt{r - 1}}{1 + \sqrt{r} \cdot \sqrt{r - 1}}) \\ = \sum_{r = 1}^{n} (\tan^{- 1} (\sqrt{r}) - \tan^{- 1} (\sqrt{r - 1})) \\ = (\tan^{- 1} (\sqrt{1}) - \tan^{- 1} (\sqrt{0}) + (\tan^{- 1} (\sqrt{2}) - \tan^{- 1} (\sqrt{1})) \\ + (\tan^{- 1} (\sqrt{3}) - \tan^{- 1} (\sqrt{2})) + (\tan^{- 1} (\sqrt{4}) - \tan^{- 1} (\sqrt{3})) \\ + \dots + (\tan^{- 1} (\sqrt{n}) - \tan^{- 1} \sqrt{(n - 1})) \\ = \tan^{- 1} (\sqrt{n}) \end{array}$