Q. If$\ \alpha + \beta + \gamma = 2\pi,\$then
A) $\tan{\frac{\alpha}{2} + \tan{\frac{\beta}{2} + \tan{\frac{\gamma}{2} = \tan{\frac{\alpha}{2}\tan{\frac{\beta}{2}\tan\frac{\gamma}{2}}}}}}$
B) $\tan{\frac{\alpha}{2}\tan{\frac{\beta}{2} + \tan{\frac{\beta}{2}\tan\frac{\gamma}{2} + \tan{\frac{\gamma}{2}\tan{\frac{\alpha}{2} = 1}}}}}$
C) $\tan{\frac{\alpha}{2} + \tan{\frac{\beta}{2} + \tan{\frac{\gamma}{2} = - \tan\frac{\alpha}{2}\tan{\frac{\beta}{2}\tan\frac{\gamma}{2}}}}}$
D) None of the above
Explanation
(a)
We have, $\alpha + \beta + \gamma = 2\pi$
$\Rightarrow \frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} = \pi$
$\Rightarrow \frac{\alpha}{2} + \frac{\beta}{2} = \pi - \frac{\gamma}{2}$
$\Rightarrow \tan{\left( \frac{\alpha}{2} + \frac{\beta}{2} \right) = \tan\left( \pi - \frac{\gamma}{2} \right)}$
$\Rightarrow \frac{\tan{\frac{\alpha}{2} + \tan\frac{\beta}{2}}}{1 - \tan{\frac{\alpha}{2}\tan\frac{\beta}{2}}} = - \tan\frac{\gamma}{2}$
$\Rightarrow \tan{\frac{\alpha}{2} + \tan{\frac{\beta}{2} + \tan{\frac{\gamma}{2} = \tan{\frac{\alpha}{2}\tan{\frac{\beta}{2}\tan\frac{\gamma}{2}}}}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.