# If tan θ

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 :

$\frac{1+\tan \theta_{1} \tan \theta_{2}}{1-\tan \theta_{1} \tan \theta_{2}}$

$=\frac{1+k}{1-k}$