Discussion Forum

If $\tan \theta_{1}=k \cot \theta_{2}$ , then $\frac{\cos (\theta_{1}+\theta_{2})}{\cos (\theta_{1}-\theta_{2}) }=$

Answer:

Explanation:

 We have $\tan \theta_{1}=k \cot \theta_{2}$

$\Rightarrow$  $\tan \theta_{1} \tan \theta_{2}$=k

Consider, $\frac{\cos (\theta_{1}+\theta_{2})}{\cos (\theta_{1}-\theta_{2})}=\frac{\cos  \theta_{1}\cos \theta_{2}-\sin  \theta_{1}\sin \theta_{2}}{\cos \theta_{1}\cos\theta_{2}+\sin \theta_{1}\sin \theta_{2}}$

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

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