2 sin 1 3 5 tan 1 24 7

Asked by admin @ in Math viewed by 349 People

Prove that 2 sin^−1 3/5=tan^−1 24/7.

Answered by admin @


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.


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