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MCQ

If $\tan \beta = \cos \theta \tan \alpha ,$ then ${\tan ^2}\frac{\theta }{2} = $

A
$\frac{{\sin (\alpha + \beta )}}{{\sin (\alpha - \beta )}}$
B
$\frac{{\cos (\alpha - \beta )}}{{\cos (\alpha + \beta )}}$
✓
$\frac{{\sin (\alpha - \beta )}}{{\sin (\alpha + \beta )}}$
D
$\frac{{\cos (\alpha + \beta )}}{{\cos (\alpha - \beta )}}$

✓

Answer

Correct option: C.

$\frac{{\sin (\alpha - \beta )}}{{\sin (\alpha + \beta )}}$

c
(c) ${\tan ^2}\frac{\theta }{2} = \frac{{1 - \cos \theta }}{{1 + \cos \theta }} $

$= \frac{{\tan \alpha - \tan \beta }}{{\tan \alpha + \tan \beta }} $

$= \frac{{\sin (\alpha - \beta )}}{{\sin (\alpha + \beta )}}$.

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