Prove
Question:

Prove $\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}$

Solution:

L.H.S. $=\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{8}$

$=\tan ^{-1}\left(\frac{\frac{1}{5}+\frac{1}{7}}{1-\frac{1}{5} \times \frac{1}{7}}\right)+\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{8}}{1-\frac{1}{3} \times \frac{1}{8}}\right) \quad\left[\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}\right]$

$=\tan ^{-1}\left(\frac{7+5}{35-1}\right)+\tan ^{-1}\left(\frac{8+3}{24-1}\right)$

$=\tan ^{-1} \frac{12}{34}+\tan ^{-1} \frac{11}{23}$

$=\tan ^{-1} \frac{6}{17}+\tan ^{-1} \frac{11}{23}$

$=\tan ^{-1}\left(\frac{\frac{6}{17}+\frac{11}{23}}{1-\frac{6}{17} \times \frac{11}{23}}\right)$

$=\tan ^{-1}\left(\frac{138+187}{391-66}\right)$

$=\tan ^{-1}\left(\frac{325}{325}\right)=\tan ^{-1} 1$

$=\frac{\pi}{4}=$ R.H.S

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