If + + = 2 , then - Vidyadip

MCQ

If $\alpha + \beta + \gamma = 2\pi ,$ then

✓
$\tan \frac{\alpha }{2} + \tan \frac{\beta }{2} + \tan \frac{\gamma }{2} = \tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2}$
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 these

✓

Answer

Correct option: A.

$\tan \frac{\alpha }{2} + \tan \frac{\beta }{2} + \tan \frac{\gamma }{2} = \tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2}$

a
(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}$

$= \tan \frac{\alpha }{2}\tan \frac{\beta }{2}\tan \frac{\gamma }{2}$.

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