Question

If $\cos (A-B)=\frac{3}{5}$ and $\tan A\tan B=2$, then?

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

Answer 1

Answer: (A), (B)

The correct options are
A $\cos A\cos B=\frac{1}{5}$
B $\cos (A+B)=-\frac{1}{5}$

Given, $\cos (A-B)=\frac{3}{5}\Rightarrow \cos A\cos B+\sin A\sin B=\frac{3}{5}---\left(1\right)$

$\tan A\tan B=2\Rightarrow 2\cos A\cos B=\sin A\sin B---\left(2\right)$ [expanding formula]

Substituting $\left(2\right)$ in $\left(1\right)$

$\cos A\cos B2\cos A\cos B=\frac{3}{5}$

$\Rightarrow \cos A\cos B=\frac{1}{5}\Rightarrow \sin A\sin B=\frac{2}{5}---\left(3\right)$

$\cos (A+B)=\cos A\cos B-\sin A\sin B=\frac{1}{5}-\frac{2}{5}=-\frac{1}{5}$

$\therefore$ Option $A,B$ are both correct