Question

Prove that: ${\sin }^{-1}\left(\frac{4}{5}\right)+2{\tan }^{-1}\left(\frac{1}{3}\right)=\frac{\pi }{2}$

Answer 1

$\begin{array}{c}Toprove{\sin }^{-1}\left(\frac{4}{5}\right)+2{\tan }^{-1}\left(\frac{1}{3}\right)=\frac{\pi }{2}\\ now,\\ L.H.S={\sin }^{-1}\left(\frac{4}{5}\right)+2{\tan }^{-1}\left(\frac{1}{3}\right)\\ ={\sin }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left[\frac{2\times \frac{1}{3}}{1-{\left(\frac{1}{3}\right)}^2}\right]\left[\therefore 2{\tan }^{-1}x=\frac{2x}{1-x^2}\right]\\ ={\sin }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left(\frac{2/3}{8/9}\right)\\ ={\sin }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left(\frac{3}{4}\right)\\ ={\tan }^{-1}\left[\frac{\frac{4}{5}}{\sqrt{1-{\left(\frac{4}{5}\right)}^2}}\right]+{\tan }^{-1}\left(\frac{3}{4}\right)\left[\therefore {\sin }^{-1}x={\tan }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\right]\\ ={\tan }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left(\frac{4}{5}\right)\\ ={\tan }^{-1}\left(\frac{4}{5}\right)+{\cot }^{-1}\left(\frac{4}{3}\right)\left[as,{\tan }^{-1}\left(\frac{1}{x}\right)={\cot }^{-1}x\right]\\ =\frac{\pi }{2}\left[\therefore {\tan }^{-}x+{\cot }^{-1}x=\frac{\pi }{2},forallx\in R\right]\\ =R.H.S\\ Hence,\\ L.H.S=R.H.Sproved\end{array}$