Question

Prove that 𝒕𝒂𝒏𝒉−𝟏(𝒔𝒊𝒏 𝜽) = 𝒄𝒐𝒔𝒉−𝟏(𝒔𝒆𝒄 𝜽) 

Answer 1

${\displaystyle L.H.S={\tanh }^1\left(\sin \theta \right)}$

We know that, ${\displaystyle {\tanh }^{-1}\left(x\right)=\frac{1}{2}\log \left(\frac{1+x}{1-x}\right)}$

∴ ${\displaystyle L.H.S=\frac{1}{2}\log \left(\frac{1+\sin \theta }{1-\sin \theta }\right)}$

${\displaystyle R.H.S={\cosh }^{-1}\left(\sec \theta \right)}$

We know that , ${\displaystyle {\cosh }^{-1}\left(x\right)=\log (x+\sqrt{x^2-1})}$

∴ R.H.S =${\displaystyle \log (\sec \theta +\sqrt{{\sec }^2\theta -1})}$

= ${\displaystyle \log (\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta })}$    ...........${\displaystyle \{\sqrt{{\sec }^2\theta -1}=\tan \theta =\frac{\sin \theta }{\cos \theta }\}}$ 

=${\displaystyle \log \left(\frac{1+\sin \theta }{\cos \theta }\right)}$

=${\displaystyle \log (1+\frac{\sin \theta }{\sqrt{1-{\sin }^2\theta }})}$

=${\displaystyle \log \left(\frac{\sqrt{1+\sin \theta }}{\sqrt{1-\sin \theta }}\right)}$

= ${\displaystyle \frac{1}{2}\log \left(\frac{1+\sin \theta }{1-\sin \theta }\right)}$

∴ ${\displaystyle {\tanh }^{-1}\left(\sin \theta \right)={\cosh }^{-1}\left(\sec \theta \right)}$

Hence Proved.