Question

Prove that 2 sin^−1 3/5=tan^−1 24/7.

Answer 1

Use the identity,

$\sin 2\theta =2\sin \theta \cos \theta\\ 2\theta = \sin ^{-1} (2\sin \theta \cos \theta)$

Here,

$LHS=2\sin ^{-1}(\frac{3}{5}) =\sin ^{-1}(2*\frac{3}{5}\sqrt{1-(\frac{3}{5})^2}) \\ LHS=2\sin ^{-1}(\frac{3}{5}) =\sin ^{-1}(2*\frac{3}{5}\frac{4}{5}) \\ LHS=2\sin ^{-1}(\frac{3}{5}) =\sin ^{-1}(\frac{24}{25}) \\ LHS=2\sin ^{-1}(\frac{3}{5}) =\tan ^{-1}(\frac{24}{\sqrt{25^2-24^2}}) \\ LHS=2\sin ^{-1}(\frac{3}{5}) =\tan ^{-1}(\frac{24}{\sqrt{49}}) \\ LHS=2\sin ^{-1}(\frac{3}{5}) =\tan ^{-1}(\frac{24}{7}) =RHS$

The proof is complete.