# Given tan x tan y = p and cos(x+y)=q .Show that sin x sin y =(pq)/(1-q) and cos(x-y)= (q(1+p))/(1-p) ?

Aug 19, 2015

The first part:

$\frac{p q}{1 - q}$

$= \frac{\tan x \tan y \cos \left(x + y\right)}{1 - \tan x \tan y}$

Using the identities $\tan x = \sin \frac{x}{\cos} x$ and $\cos \left(x + y\right) = \cos x \cos y - \sin x \sin y$ on the numerator and denominator, and cross-multiplying in the denominator:

$= \frac{\left\{\frac{\sin x \sin y}{\cancel{\cos x \cos y}}\right\} \cancel{\left\{\cos x \cos y - \sin x \sin y\right\}}}{\frac{\cancel{\cos x \cos y - \sin x \sin y}}{\cancel{\cos x \cos y}}}$

$= \sin x \sin y$

The second part:

$\frac{q \left(1 + p\right)}{1 - p}$

$= \left(\cos x \cos y - \sin x \sin y\right) \left[\frac{1 + \tan x \tan y}{1 - \tan x \tan y}\right]$

From here we can do the following:

$\frac{1 + \tan x \tan y}{1 - \tan x \tan y} = \frac{1 + \frac{\sin x \sin y}{\cos x \cos y}}{1 - \frac{\sin x \sin y}{\cos x \cos y}}$

Cross-multiply to get:

$= \frac{\frac{\left(\cos x \cos y\right) + \left(\sin x \sin y\right)}{\cancel{\cos x \cos y}}}{\frac{\left(\cos x \cos y\right) - \left(\sin x \sin y\right)}{\cancel{\cos x \cos y}}}$

$= \frac{\cos x \cos y + \sin x \sin y}{\cos x \cos y - \sin x \sin y}$

Thus, we have:

$= \cancel{\left(\cos x \cos y - \sin x \sin y\right)} \left[\frac{\cos x \cos y + \sin x \sin y}{\cancel{\left(\cos x \cos y - \sin x \sin y\right)}}\right]$

$= \cos x \cos y + \sin x \sin y$

$= \cos \left(x - y\right)$