Question

If $\cos (A-B)=3/5$ and $\tan A\tan B=2$, then which one of the following is true ?

• A. $\sin (A+B)=\frac{1}{5}$
• B. $\sin (A+B)=-\frac{1}{5}$
• C. $\cos (A-B)=\frac{1}{5}$
• D. $\cos (A+B)=-\frac{1}{5}$

Answer 1

Answer: (D)

Given, $\cos (A-B)=\frac{3}{5}$ and $\tan A\tan B=2$
$\Rightarrow \frac{\sin A\sin B}{\cos A\cos B}=2$
Using componendo and dividendo
$\frac{\sin A\sin B+\cos A\cos B}{\sin A\sin B-\cos A\cos B}=\frac{2+1}{2-1}$
$\Rightarrow \frac{\cos (A-B)}{-\cos (A+B)}=\frac{3}{1}$
$\Rightarrow \frac{3/5}{-\cos (A+B)}=\frac{3}{1}$
$\Rightarrow -3\cos (A+B)=\frac{3}{5}$
$\Rightarrow \cos (A+B)=-\frac{1}{5}$