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  4. The value of tan -1 (4/7)+ tan -1 (4/19)+ tan -1 (4/39)+ tan -1 (4/67)+ ldots ldots . . ∞ equals

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Q. The value of $\tan ^{-1} \frac{4}{7}+\tan ^{-1} \frac{4}{19}+\tan ^{-1} \frac{4}{39}+\tan ^{-1} \frac{4}{67}+\ldots \ldots . . \infty$ equals

Inverse Trigonometric Functions

Solution:

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$=7+\frac{(n-1)}{2}[24+8(n-2)]=4 n^2+3 $
$\therefore T_n^{\prime}=\tan ^{-1} \frac{4}{4 n^2+3}=\tan ^{-1} \frac{1}{n^2+\frac{3}{4}}=\tan ^{-1} \frac{1}{1+\left(n^2-\frac{1}{4}\right)} $
$=\tan ^{-1}\left[\frac{\left(n+\frac{1}{2}\right)-\left(n-\frac{1}{2}\right)}{1+\left(n+\frac{1}{2}\right)\left(n-\frac{1}{2}\right)}\right]=\tan ^{-1}\left(n+\frac{1}{2}\right)-\tan ^{-1}\left(n-\frac{1}{2}\right) $
$\text { Hence } S_{\infty}=\displaystyle\sum_{n=1}^{\infty} T_n^{\prime}=\frac{\pi}{2}-\tan ^{-1} \frac{1}{2}=\tan ^{-1} 1+\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}-\tan ^{-1} \frac{1}{2}=\tan ^{-1} 1+\cot ^{-1} 3$