Question

$tan\alpha -tan\left(\beta /2\right)+\sqrt{3}tan\left(\gamma /4\right)$ is equal to

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

The correct option is D 1

$\alpha =ta{n}^{-1}\left(\frac{1}{2}\right)+ta{n}^{-1}\left(\frac{1}{3}\right)=ta{n}^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\times \frac{1}{3}}\right)=ta{n}^{-1}\left(1\right)=\frac{\pi }{4}\beta =co{s}^{-1}\frac{2}{3}+co{s}^{-1}\frac{\sqrt{5}}{3}=co{s}^{-1}\frac{2}{3}+si{n}^{-1}\frac{2}{3}=\frac{\pi }{2}\gamma =si{n}^{-1}\left(sin\frac{2\pi }{3}\right)+\frac{1}{2}co{s}^{-1}\left(cos\frac{2\pi }{3}\right)=si{n}^{-1}\left(sin(\pi -\frac{\pi }{3})\right)+\frac{1}{2}co{s}^{-1}\left(cos\frac{2\pi }{3}\right)=\frac{\pi }{3}+\frac{1}{2}\times \frac{2\pi }{3}=\frac{2\pi }{3}$
Therefore,
$tan\alpha -tan\frac{\beta }{2}+\sqrt{3}tan\frac{\gamma }{4}=tan\frac{\pi }{4}-tan\frac{\pi }{4}+\sqrt{3}tan\frac{2\pi }{3\times 4}=1-1+\sqrt{3}tan\frac{\pi }{6}=\sqrt{3}\times \frac{1}{\sqrt{3}}=1$