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If $\tan \theta_1=k \cot \theta_2$, then $\frac{\cos \left(\theta_1+\theta_2\right)}{\cos \left(\theta_1-\theta_2\right)}=$
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Verified Answer
The correct answer is:
$\frac{1-k}{1+k}$
We have, $\tan \theta_1=k \cot \theta_2$
$\Rightarrow \quad \tan \theta_1 \tan \theta_2=k$
Consider, $\frac{\cos \left(\theta_1+\theta_2\right)}{\cos \left(\theta_1-\theta_2\right)}=\frac{\cos \theta_1 \cos \theta_2-\sin \theta_1 \sin \theta_2}{\cos \theta_1 \cos \theta_2+\sin \theta_1 \sin \theta_2}$
$$
=\frac{1-\tan \theta_1 \tan \theta_2}{1+\tan \theta_1 \tan \theta_2}
$$
$$
=\frac{1-k}{1+k}
$$
$\Rightarrow \quad \tan \theta_1 \tan \theta_2=k$
Consider, $\frac{\cos \left(\theta_1+\theta_2\right)}{\cos \left(\theta_1-\theta_2\right)}=\frac{\cos \theta_1 \cos \theta_2-\sin \theta_1 \sin \theta_2}{\cos \theta_1 \cos \theta_2+\sin \theta_1 \sin \theta_2}$
$$
=\frac{1-\tan \theta_1 \tan \theta_2}{1+\tan \theta_1 \tan \theta_2}
$$
$$
=\frac{1-k}{1+k}
$$
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