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Mathematics Question on Inverse Trigonometric Functions

cot⁡−1(2.12)+cot⁡−1(2.22)+cot⁡−1(2.32)+\cot^{-1} (2.1^{2})+\cot^{-1} (2.2^{2})+\cot^{-1}(2.3^2)+.........up to ∞\infty=

A

π5\frac {\pi}{5}

B

π4\frac {\pi}{4}

C

π3\frac {\pi}{3}

D

π2\frac {\pi}{2}

Answer

π4\frac {\pi}{4}

Explanation

Solution

cot⁡−1(2⋅12)+cot⁡−1(2⋅22)+cot⁡−1(2⋅32)+…∞\cot ^{-1}\left(2 \cdot 1^{2}\right)+\cot ^{-1}\left(2 \cdot 2^{2}\right)+\cot ^{-1}\left(2 \cdot 3^{2}\right)+\ldots \infty
=∑r=1∞cot⁡−1(2⋅r2)=∑r=1∞tan⁡−1(12r2)=\sum\limits_{r=1}^{\infty} \cot ^{-1}\left(2 \cdot r^{2}\right)=\sum\limits_{r=1}^{\infty} \tan ^{-1}\left(\frac{1}{2 r^{2}}\right)
=∑r=1∞tan⁡−1((1+2r)+(1−2r)1−(1+2r)(1−2r))=\sum\limits_{r=1}^{\infty} \tan ^{-1}\left(\frac{(1+2 r)+(1-2 r)}{1-(1+2 r)(1-2 r)}\right)
=∑r=1∞[tan⁡−1(1+2r)+tan⁡−1(1−2r)]=\sum\limits_{r=1}^{\infty}\left[\tan ^{-1}(1+2 r)+\tan ^{-1}(1-2 r)\right]
=tan⁡−13−tan⁡−11+tan⁡−15−tan⁡−13=\tan ^{-1} 3-\tan ^{-1} 1+\tan ^{-1} 5-\tan ^{-1} 3
+tan⁡−17−tan⁡−15+…+tan⁡−1∞+\tan ^{-1} 7-\tan ^{-1} 5+\ldots+\tan ^{-1} \infty
=−π4+π2=π4=-\frac{\pi}{4}+\frac{\pi}{2}=\frac{\pi}{4}