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  4. The value of tan ( displaystyle∑r=1∞ tan -1((4/4 r2+3))) is equal to

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Q. The value of $\tan \left(\displaystyle\sum_{r=1}^{\infty} \tan ^{-1}\left(\frac{4}{4 r^2+3}\right)\right)$ is equal to

Inverse Trigonometric Functions

Solution:

$\tan ^{-1}\left(\frac{4}{4 r^2+3}\right)=\tan ^{-1}\left(\frac{1}{r^2+\frac{3}{4}}\right)=\tan ^{-1}\left(\frac{1}{1+\left(r+\frac{1}{2}\right)\left(r-\frac{1}{2}\right)}\right)=\tan ^{-1}\left(r+\frac{1}{2}\right)-\tan ^{-1}\left(r-\frac{1}{2}\right)$
$\therefore \displaystyle\sum_{r=1}^n \tan ^{-1}\left(\frac{4}{4 r^2+3}\right)=\left(\tan ^{-1}\left(n+\frac{1}{2}\right)-\tan ^{-1} \frac{1}{2}\right) $
$\Rightarrow \displaystyle\sum_{r=1}^{\infty} \tan ^{-1}\left(\frac{4}{4 r^2+3}\right)=\cot ^{-1} \frac{1}{2}=\left(\tan ^{-1} 2\right) $
$\therefore \tan \left(\tan ^{-1} 2\right)=2 $