Question

If $\cos \left(A+B\right)=\frac{3}{5}$ and $\tan A\tan B=2$, then $\cos A\cos B$

• A. $\frac{2}{5}$
• B. $\frac{1}{5}$
• C. $\frac{3}{5}$
• D. $-\frac{3}{5}$

Answer 1

Answer: (D)

$-\frac{3}{5}$

We have,

$\cos (A+B)=\frac{3}{5}$ ……. (1)

$\tan A\tan B=2$ ……. (2)

Since,

$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

$\tan (A+B)=\frac{\tan A+\tan B}{1-2}$

$\tan (A+B)=-(\tan A+\tan B)$

$\frac{\sin (A+B)}{\cos (A+B)}=-(\frac{\sin A}{\cos A}+\frac{\sin B}{\cos B})$

$\frac{\sin (A+B)}{\cos (A+B)}=-\left(\frac{\sin A\cos B+\cos A\sin B}{\cos A\cos B}\right)$

$\frac{\sin (A+B)}{\cos (A+B)}=-\left(\frac{\sin (A+B)}{\cos A\cos B}\right)$

$\frac{1}{\cos (A+B)}=-\left(\frac{1}{\cos A\cos B}\right)$

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

From equation $\left(1\right)$, we get

$\frac{3}{5}=-\cos A\cos B$

$\cos A\cos B=-\frac{3}{5}$

Hence, this is the answer.