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If $\cos (A-B)=3 / 5$ and $\tan A \tan B=2$, then which one of the following is true?
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Verified Answer
The correct answer is:
$\cos (A+B)=-\frac{1}{5}$
Given, $\cos (A-B)=\frac{3}{5}$ and $\tan A \tan B=2$ $\Rightarrow \quad \frac{\sin A \sin B}{\cos A \cos B}=2$
Using componendo and dividendo
$\begin{aligned} & \frac{\sin A \sin B+\cos A \cos B}{\sin A \sin B-\cos A \cos B}=\frac{2+1}{2-1} \\ \Rightarrow & \frac{\cos (A-B)}{-\cos (A+B)}=\frac{3}{1} \\ \Rightarrow & \frac{3 / 5}{-\cos (A+B)}=\frac{3}{1} \\ \Rightarrow & -3 \cos (A+B)=\frac{3}{5} \\ \Rightarrow & \cos (A+B)=-\frac{1}{5}\end{aligned}$
Using componendo and dividendo
$\begin{aligned} & \frac{\sin A \sin B+\cos A \cos B}{\sin A \sin B-\cos A \cos B}=\frac{2+1}{2-1} \\ \Rightarrow & \frac{\cos (A-B)}{-\cos (A+B)}=\frac{3}{1} \\ \Rightarrow & \frac{3 / 5}{-\cos (A+B)}=\frac{3}{1} \\ \Rightarrow & -3 \cos (A+B)=\frac{3}{5} \\ \Rightarrow & \cos (A+B)=-\frac{1}{5}\end{aligned}$
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