Question

Let $\tan \left(\frac{1}{2}{\cos }^{-1}\frac{\sqrt{5}}{3}\right)=\frac{a}{b+\sqrt{c}},$ where $a,b,c$ are co primes. Then the value of $a+b+c$ is

Answer 1

Let ${\cos }^{-1}\frac{\sqrt{5}}{3}=\theta$
Then $\cos \theta =\frac{\sqrt{5}}{3}$ and $\sin \theta =\frac{\sqrt{9-5}}{3}=\frac{2}{3}$

$\tan \left(\frac{1}{2}{\cos }^{-1}\frac{\sqrt{5}}{3}\right)$
$=\tan \frac{\theta }{2}$
$=\frac{\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}}=\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\cos }^2\frac{\theta }{2}}=\frac{\sin \theta }{1+\cos \theta }$
$=\frac{2/3}{1+\left(\sqrt{5}/3\right)}$
$=\frac{2}{3+\sqrt{5}}$
Hence, $a=2,b=3$ and $c=5$
$a+b+c=10$