Question

If $3\tan \theta \tan \alpha =1$, then find the value of $\frac{\cos (\theta -\alpha )}{\cos (\theta +\alpha )}=$

• A. $1$
• B. $-1$
• C. $2$
• D. $-2$

Answer 1

The correct option is C $2$

$\text{solution:}3\tan \theta \cdot \tan \alpha =1$

$\Rightarrow 3\sin \theta \sin \alpha =\cos \theta \cos \alpha$

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

$=\frac{3\sin \theta \sin \alpha +\sin \theta \sin \alpha }{3\sin \theta \sin \alpha -\sin \theta \sin \alpha }$

$=\frac{4\sin \theta \sin \alpha }{2\sin \theta \sin \alpha }=2$

$\text{Answer: option (c)}$