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JEE Main 2024
Inverse Trigonometric Functions
Inverse Trigonometric Functions
Hard

Question

The sum of the infinite series cot1(74)+cot1(194)+cot1(394)+cot1(674)+\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. is :

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Solution

cot1(74)+cot1(194)+cot1(394)+cot1(674)+Tr=cot1(4r2+34)Tr=tan1(1(34+r2))\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} Tr=tan1((r+12)(r12)1+r21/4)Tr=tan1((r+12)(r12)1+(r+12)(r12))Tr=tan1(r+12)tan1(r12)T1=tan1(32)tan1(12)\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} T2=tan1(52)tan1(32)Tn=tan1(2n+12)tan1(12)ΣTr=tan1(2n+12)tan1(12)ΣTr=π2tan1(12)\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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