If $\frac{\sin(x+y)}{\sin(x-y)} = \frac{a+b}{a-b}$,then $\frac{\tan x}{\tan y} = $

In an isosceles right-angled triangle,a straight line is drawn from the midpoint of one of the equal sides to the opposite vertex. Then a pair of possible values of the cotangents of the two angles so formed at that vertex are

Prove that $\cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)$

Prove that $\frac{\sin (x+y)}{\sin (x-y)} = \frac{\tan x + \tan y}{\tan x - \tan y}$.

$\cos 20^{\circ} + \cos 30^{\circ} + \cos 40^{\circ} = $

Prove that: $(\cos x+\cos y)^{2}+(\sin x-\sin y)^{2}=4 \cos ^{2} \frac{x+y}{2}$