∑r=1n sin−1⁡r−r−1r(r+1)=

∑r=1n sin−1⁡r−r−1r(r+1)=∑r=1n tan−1⁡r−r−11+rr−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⁡n−tan−1⁡n−1=tan−1⁡n−tan−1⁡(0)=tan−1⁡n

∑r=1n sin−1⁡r−r−1r(r+1)=

∑r=1n sin−1⁡r−r−1r(r+1)=∑r=1n tan−1⁡r−r−11+rr−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⁡n−tan−1⁡n−1=tan−1⁡n−tan−1⁡(0)=tan−1⁡n

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

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

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

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

$\tan^{- 1} (\sqrt{n}) - \frac{π}{4}$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

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