JEE MAIN - Mathematics (2025 - 4th April Evening Shift - No. 15)
Explanation
$$\begin{aligned} & \cot ^{-1}\left(\frac{7}{4}\right)+\cot ^{-1}\left(\frac{19}{4}\right)+\cot ^{-1}\left(\frac{39}{4}\right)+\cot ^{-1}\left(\frac{67}{4}\right)+\ldots \\ & T_r=\cot ^{-1}\left(\frac{4 r^2+3}{4}\right) \\ & T_r=\tan ^{-1}\left(\frac{1}{\left(\frac{3}{4}+r^2\right)}\right) \end{aligned}$$
$$\begin{aligned} & T_r=\tan ^{-1}\left(\frac{\left(r+\frac{1}{2}\right)-\left(r-\frac{1}{2}\right)}{1+r^2-1 / 4}\right) \\ & T_r=\tan ^{-1}\left(\frac{\left(r+\frac{1}{2}\right)-\left(r-\frac{1}{2}\right)}{1+\left(r+\frac{1}{2}\right)\left(r-\frac{1}{2}\right)}\right) \\ & T_r=\tan ^{-1}\left(r+\frac{1}{2}\right)-\tan ^{-1}\left(r-\frac{1}{2}\right) \\ & T_1=\tan ^{-1}\left(\frac{3}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right) \end{aligned}$$
$$\begin{aligned} & T_2=\tan ^{-1}\left(\frac{5}{2}\right)-\tan ^{-1}\left(\frac{3}{2}\right) \\ & \vdots \qquad\qquad \vdots \qquad \qquad\qquad \vdots \\ & T_n=\tan ^{-1}\left(\frac{2 n+1}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right) \\ & \Sigma T_r=\tan ^{-1}\left(\frac{2 n+1}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right) \\ & \Sigma T_r=\frac{\pi}{2}-\tan ^{-1}\left(\frac{1}{2}\right) \end{aligned}$$
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