Question

If $\tan x\tan y=\frac{1}{2}$ and $\cos (x-y)=p\cos (x+y)$, then the value of $p$ is

Answer 1

We know that,
$\tan x\tan y=\frac{1}{2}$
$\cos (x-y)=p\cos (x+y)\Rightarrow \cos x\cos y+\sin x\sin y=p(\cos x\cos y-\sin x\sin y)$
Dividing both sides by $\cos x\cos y$,
$\Rightarrow 1+\tan x\tan y=p(1-\tan x\tan y)$
Putting the value of $\tan x\tan y=\frac{1}{2}$
$\frac{3}{2}=\frac{p}{2}\therefore p=3$