Question

If $\alpha ={\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3},$ $\beta ={\cos }^{-1}\frac{2}{3}+{\cos }^{-1}\frac{\sqrt{5}}{3}$ and $\gamma ={\sin }^{-1}\left(\sin \frac{2\pi }{3}\right)+\frac{1}{2}{\cos }^{-1}\left(\cos \frac{2\pi }{3}\right),$ then the value of $\tan \alpha -\tan \left(\frac{\beta }{2}\right)+\sqrt{3}\tan \left(\frac{\gamma }{4}\right)$ is equal to

• A. $2$
• B. $1$
• C. $0$
• D. $-1$

Answer 1

The correct option is B $1$

$\alpha ={\tan }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{3}$
$={\tan }^{-1}\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}={\tan }^{-1}1=\frac{\pi }{4}$

$\beta ={\cos }^{-1}\frac{2}{3}+{\cos }^{-1}\frac{\sqrt{5}}{3}={\tan }^{-1}\frac{\sqrt{5}}{2}+{\tan }^{-1}\frac{2}{\sqrt{5}}={\tan }^{-1}\frac{\sqrt{5}}{2}+{\cot }^{-1}\frac{\sqrt{5}}{2}=\frac{\pi }{2}$

$\gamma ={\sin }^{-1}\left(\sin \frac{2\pi }{3}\right)+\frac{1}{2}{\cos }^{-1}\left(\cos \frac{2\pi }{3}\right)={\sin }^{-1}\left(\sin \frac{\pi }{3}\right)+\frac{1}{2}\times \frac{2\pi }{3}=\frac{\pi }{3}+\frac{\pi }{3}=\frac{2\pi }{3}$

Now, $\tan \alpha -\tan \left(\frac{\beta }{2}\right)+\sqrt{3}\tan \left(\frac{\gamma }{4}\right)$
$=\tan \frac{\pi }{4}-\tan \frac{\pi }{4}+\sqrt{3}\tan \frac{\pi }{6}$
$=1-1+1=1$