STD 12 - 2. inverse trigonometric function — Mathematics STD 12 Science — Question

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

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

But $\cos 2\theta = \frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }} $

$\Rightarrow \frac{{\sqrt 5 }}{3} = \frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$

==>$\sqrt 5 + \sqrt 5 {\tan ^2}\theta = 3 - 3{\tan ^2}\theta $

==> $(\sqrt 5 + 3){\tan ^2}\theta = 3 - \sqrt 5 $

$\Rightarrow {\tan ^2}\theta = \frac{{3 - \sqrt 5 }}{{3 + \sqrt 5 }}$

==> ${\tan ^2}\theta = \frac{{{{(3 - \sqrt 5 )}^2}}}{4} $

$\Rightarrow \tan \theta = \frac{{3 - \sqrt 5 }}{2}$

On rationalising

==> $\tan \theta = \frac{{3 - \sqrt 5 }}{2} \times \frac{{3 + \sqrt 5 }}{{3 + \sqrt 5 }} = \frac{2}{{3 + \sqrt 5 }}$.