Question

If $\tan \beta =$ cos $\theta$ tan $\alpha$, then $\sin (\alpha -\beta )=$

• A. ${\tan }^2\frac{\theta }{2}\sin (\alpha +\beta )$
• B. ${\cot }^2\frac{\theta }{2}\sin (\alpha +\beta )$
• C. ${\tan }^2\theta \sin (\alpha +\beta )$
• D. ${\cot }^2\theta \sin (\alpha +\beta )$

Answer 1

The correct option is A ${\tan }^2\frac{\theta }{2}\sin (\alpha +\beta )$

Given $tan\beta =cos\theta tan\alpha ......1$

$sin(\alpha -\beta )=sin\alpha cos\beta -sin\beta cos\alpha$

from $1$

$sin\beta cos\alpha =cos\theta sin\alpha cos\beta$

$\Rightarrow sin(\alpha -\beta )=cos\beta sin\alpha (1-cos\theta )=2si{n}^2\frac{\theta }{2}cos\beta sin\alpha$

$\sin (\alpha +\beta )=\cos \beta \sin \alpha (1+\cos \theta )$

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

$=\frac{2si{n}^2\frac{\theta }{2}cos\beta sin\alpha }{(1+cos\theta )cos\beta sin\alpha }$

$\frac{sin(\alpha -\beta )}{sin(\alpha +\beta )}=ta{n}^2\frac{\theta }{2}$

$\Rightarrow sin(\alpha -\beta )=ta{n}^2\frac{\theta }{2}\left(sin\right(\alpha +\beta \left)\right)$