Question

If $\cot A\cot B=2,\cos (A+B)=3/5$, then $\sin A.\sin B$ is equal to

• A. $\frac{2}{5}$
• B. $\frac{1}{5}$
• C. $\frac{4}{5}$
• D. $\frac{3}{5}$

Answer 1

Answer: (D)

$\frac{3}{5}$

Given $\Rightarrow \cot A\cot B=2$
So, $\cos A\cos B=2\sin A\sin B$ ...(1)
Given $\cos (A+B)=\frac{3}{5}$
So, $\cos A\cos B-\sin A\sin B=\frac{3}{5}$ ....(2)
From eq.(1) and eq.(2)
$\sin A\sin B=\frac{3}{5}$