Question:
If $\tan \theta_{1} \tan \theta_{2}=k$, then $\frac{\cos \left(\theta_{1}-\theta_{2}\right)}{\cos \left(\theta_{1}+\theta_{2}\right)}=$
(a) $\frac{1+k}{1-k}$
(b) $\frac{1-k}{1+k}$
(c) $\frac{k+1}{k-1}$
(d) $\frac{k-1}{k+1}$
Solution:
(a) $\frac{1+k}{1-k}$
$\frac{\cos \left(\theta_{1}-\theta_{2}\right)}{\cos \left(\theta_{1}+\theta_{2}\right)}$
$=\frac{\cos \theta_{1} \cos \theta_{2}+\sin \theta_{1} \sin \theta_{2}}{\cos \theta_{1} \cos \theta_{2}-\sin \theta_{1} \sin \theta_{2}}$
Dividing numerator and denominator by $\cos \theta_{1} \cos \theta_{2}$, we get :
$\frac{1+\tan \theta_{1} \tan \theta_{2}}{1-\tan \theta_{1} \tan \theta_{2}}$
$=\frac{1+k}{1-k}$