Question
Prove that:
$2\sin^{-1}\frac{3}{5}=\tan^{-1}\frac{24}{7}$
$2\sin^{-1}\frac{3}{5}=\tan^{-1}\frac{24}{7}$
$=\tan^{-1}\Bigg(\frac{2\times\frac{3}{4}}{1-\left(\frac{3}{4}\right)^2}\Bigg)$ $\bigg[2\tan^{-1}x=\tan^{-1}\frac{2x}{1-x^2}\bigg]$
$=\tan^{-1}\Bigg(\frac{\frac{3}{2}}{\frac{16-9}{16}}\Bigg)=\tan^{-1}\bigg(\frac{3}{2}\times\frac{16}{7}\bigg)$
$=\tan^{-1}\frac{24}{7}=\text{R.H.S.}$
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