Question

If $\cos (A-B)=3/5$ and $\tan A\tan B=2$, then

• A. $\cos A\cos B=1/5$
• B. $\sin A\sin B=-2/5$
• C. $\cos A\cos B=-1/5$
• D. $\sin A\sin B=-1/5$

Answer 1

The correct option is A

$\cos A\cos B=1/5$

Explanation for the correct option:

Step 1. Given that $\cos (A-B)=3/5$

$\Rightarrow$ $\begin{array}{rcl}\cos (A–B)& =& 3/5\dots \left(i\right)\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\tan A\tan B& =& 2\end{array}$

$\Rightarrow$$\begin{array}{rcl}(\sin A/\cos A).(\sin B/\cos B)& =& 2/1\end{array}$

Step 2. Using componendo and dividendo rule,

$\begin{array}{rcl}\frac{\sin A\sin B+\cos A\cos B}{\sin A\sin B–\cos A\cos B}& =& \frac{2+1}{2-1}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{\cos (A–B)}{-\cos (A+B)}& =& 3\end{array}$

$\Rightarrow$ $\begin{array}{rcl}-3/5\cos (A+B)& =& 3[From(i\left)\right]\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\cos (A+B)& =& -1/5\dots .\left(ii\right)\end{array}$

Step 3. Adding (i) and (ii),

$\begin{array}{rcl}\cos (A–B)+\cos (A+B)& =& (3/5)–(1/5)\end{array}$

$\begin{array}{rcl}2\cos A\cos B& =& 2/5\end{array}$

$\Rightarrow \cos A\cos B=\frac{1}{5}$

The correct answer is option A