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

Answer 1

The correct option is A $\frac{1+k}{1-k}$

$\tan {\theta }_1\tan {\theta }_2=k$

$\sin {\theta }_1\sin {\theta }_2=k\cos {\theta }_1\cos {\theta }_2⟶\left(1\right)$

$\because \tan \theta =\frac{\sin \theta }{\cos \theta }$

$\cos (A-B)=\cos A\cos B+\sin A\sin B$

$\cos (A-B)=\cos A\cos B-\sin A\sin B$

So,

$\frac{\cos ({\theta }_1-{\theta }_2)}{\cos ({\theta }_1+{\theta }_2)}=\frac{\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2}{\cos {\theta }_1\cos {\theta }_2-\sin {\theta }_1\sin {\theta }_2}$

Using equation (1):

$=\frac{\cos {\theta }_1\cos {\theta }_2+k\cos {\theta }_1\cos {\theta }_2}{\cos {\theta }_1\cos {\theta }_2-k\cos {\theta }_1\cos {\theta }_2}$

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