Prove that:
$\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}$
To Prove: $\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}$
Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where $x y<1$
Proof:
$\mathrm{LHS}=\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}$
$=\tan ^{-1} \frac{1}{2}+\tan ^{-1}\left(\frac{\frac{1}{5}+\frac{1}{8}}{1-\left(\frac{1}{5} \times \frac{1}{8}\right)}\right)$
$=\tan ^{-1} \frac{1}{2}+\tan ^{-1}\left(\frac{8+5}{40-1}\right)$
$=\tan ^{-1} \frac{1}{2}+\tan ^{-1}\left(\frac{13}{39}\right)$
$=\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{3}$
$=\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\left(\frac{1}{2} \times \frac{1}{3}\right)}\right)$
$=\tan ^{-1}\left(\frac{3+2}{6-1}\right)$
$=\tan ^{-1} 1$
$=\frac{\pi}{4}$
$=\mathrm{RHS}$
Therefore $\mathrm{LHS}=\mathrm{RHS}$
Hence proved.
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