MCQ
$cot^{-1}\left(x^2+\frac{3}{4}\right)+cot^{-1}\left(2^2+\frac{3}{4}\right)+cot^{-1}\left(3^2+\frac{3}{4}\right)=...........$
- A$\frac{\pi }{4}$
- ✓${{\tan }^{-1}}2$
- C${{\tan }^{-1}}3$
- Dએક પણ નહીં.
$T_n = \cot^{-1} \left( n^2 +\frac{3}{4}\right)$
$= \tan^{-1} \left(\frac{4}{4n^2+3}\right)$
$= \tan^{-1} \left(\frac{(n+\frac{1}{2}- (n-\frac{1}{2}}{1+ (n+\frac{1}{2})(n-\frac{1}{2})}\right)$
$= \tan^{-1} \left[n+\frac{1}{2}\right]- \tan^{-1} \left[n-\frac{1}{2}\right]$
$S_\infty = \tan^{-1} \infty - \tan^{-1} \frac {1}{2}$
$= \frac {\pi}{2} - \tan^{-1} \frac {1}{2}$
$= \tan^{-1} 2$
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