Question
Prove that $2 \tan ^{-1}\left(\frac{3}{4}\right)=\tan ^{-1}\left(\frac{24}{7}\right)$

Answer

$2 \tan ^{-1}\left(\frac{3}{4}\right)$
$=\tan ^{-1}\left(\frac{3}{4}\right)+\tan ^{-1}\left(\frac{3}{4}\right)$
$=\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{3}{4}}{1-\frac{3}{4} \times \frac{3}{4}}\right) \ldots \ldots . .\left[\tan ^{-1} a +\tan ^{-1} b =\tan ^{-1}\left(\frac{ a + b }{1- ab }\right)\right]$
$=\tan ^{-1}\left(\frac{\frac{6}{4}}{\frac{7}{16}}\right)$
$=\tan ^{-1}\left(\frac{24}{7}\right)$

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