Question

Prove that
$\cos h^{-1}\left(\sqrt{1+x^2}\right)=\sin h^{-1}x={\tan }^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$

Answer 1

Let $\cos h^{-1}\left(\sqrt{1+x^2}\right)=y$ .........$\left(1\right)$
$\therefore \left(\sqrt{1+x^2}\right)=\cos hy$
Squaring both sides, we get
$(1+x^2)=\cos h^{2}y$
$\Rightarrow x^2=\cos h^{2}y-1$
We know that $\cos h^{2}y-\sin h^{2}y=1$
$\Rightarrow x^2=\sin h^{2}y$
$\therefore x=\sin hy$
or $y=\sin h^{-1}x$ ..................$\left(2\right)$
Now $\frac{x}{\sqrt{1+x^2}}=\frac{sinhy}{coshy}=\tan hy$
$\therefore y=tan{h}^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$ ..............$\left(3\right)$
From eqns$\left(1\right),\left(2\right)$ and $\left(3\right)$ we have
$\cos h^{-1}\left(\sqrt{1+x^2}\right)=\sin h^{-1}x={\tan }^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$
Hence proved.