Question

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

• A. $\frac{3+\sqrt{5}}{2}$
• B. $3+\sqrt{5}$
• C. $\frac{1}{2}(3-\sqrt{5})$
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{2}(3-\sqrt{5})$

Let ${\cos }^{-1}\left(\frac{\sqrt{5}}{3}\right)=\alpha$ then $\cos \alpha =\frac{\sqrt{5}}{3}$, where $0<\alpha <\frac{\pi }{2}$.
Now, $\tan \frac{\alpha }{2}=\sqrt{\frac{1-\cos \alpha }{1+\cos \alpha }}=\sqrt{\frac{1-\sqrt{5}/3}{1+\sqrt{5}/3}}$
$=\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}=\sqrt{\frac{{(3-\sqrt{5})}^2}{9-5}}$

$=\frac{1}{2}(3-\sqrt{5})$