Question

Find the value of $\tan \left(\frac{1}{2}{\cos }^{-1}\frac{\sqrt{5}}{3}\right)$

Answer 1

To simplify:

$\left[\tan \left(\frac{1}{2}{\cos }^{-1}\frac{\sqrt{5}}{3}\right)\right]$

Let $x=\frac{1}{2}{\cos }^{-1}\frac{\sqrt{5}}{3}$

Then $\tan x=?$ ……… (1)

$x=\frac{1}{2}{\cos }^{-1}\frac{\sqrt{5}}{3}$

$2x={\cos }^{-1}\frac{\sqrt{5}}{3}$

$\Rightarrow \cos 2x=\frac{\sqrt{5}}{3}$

$\Rightarrow 2{\cos }^{2}x-1=\frac{\sqrt{5}}{3}$

$\Rightarrow 2{\cos }^{2}x=\frac{\sqrt{5}}{3}+1$

$\Rightarrow 2{\cos }^{2}x=\frac{\sqrt{5}+3}{3}$

$\Rightarrow {\cos }^{2}x=\frac{\sqrt{5}+3}{6}$

$\Rightarrow \cos x=\sqrt{\frac{\sqrt{5}+3}{6}}$

$\Rightarrow \cos x=\frac{\sqrt{\sqrt{5}+3}}{\sqrt{6}}=\frac{Base}{Hypotenouse}$

But,

$\tan x=\frac{prependicular}{base}$

By Pythagoras theorem,

$Hypotenous{e}^2=Bas{e}^2+Prependicula{r}^2$

$\tan x=\frac{\sqrt{3-\sqrt{5}}}{\sqrt{\sqrt{5}+3}}$

$x={\tan }^{-1}\frac{\sqrt{3-\sqrt{5}}}{\sqrt{\sqrt{5}+3}}$

Put in equation (1).

$={tan(tan}^{-1}\frac{\sqrt{3-\sqrt{5}}}{\sqrt{\sqrt{5}+3}})$

$=\frac{\sqrt{3-\sqrt{5}}}{\sqrt{\sqrt{5}+3}}\times \frac{\sqrt{3-\sqrt{5}}}{\sqrt{3-\sqrt{5}}}$

$=\frac{3-\sqrt{5}}{\sqrt{9-5}}$

$=\frac{3-\sqrt{5}}{\sqrt{4}}$

$=\frac{3-\sqrt{5}}{2}$

Hence, this is the answer.