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If $\alpha + \beta + \gamma = 2\pi ,$ then
$\tan \frac{\alpha }{2} + \tan \frac{\beta }{2} + \tan \frac{\gamma }{2} = \tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2}$
$\tan \frac{\alpha }{2}\tan \frac{\beta }{2} + \tan \frac{\beta }{2}\tan \frac{\gamma }{2} + \tan \frac{\gamma }{2}\tan \frac{\alpha }{2} = 1$
$\tan \frac{\alpha }{2} + \tan \frac{\beta }{2} + \tan \frac{\gamma }{2} = - \tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2}$
None of these
Solution
(a) We have $\alpha + \beta + \gamma = 2\pi $
$\Rightarrow \frac{\alpha }{2} + \frac{\beta }{2} + \frac{\gamma }{2} = \pi $
$ \Rightarrow \tan \left( {\frac{\alpha }{2} + \frac{\beta }{2} + \frac{\gamma }{2}} \right) = \tan \pi = 0$
$ \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} = 0$
$ \Rightarrow \tan \frac{\alpha }{2} + \tan \frac{\beta }{2} + \tan \frac{\gamma }{2}$
$= \tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2}$.
