Answer
$\sin (x+y)= \sin x \cos y+\cos x \sin y\\ \cos (x+y)= \cos x \cos y- \sin x \sin y\\ \sin (x-y)= \sin x \cos y -\cos x \sin y \\ \cos (x-y)= \cos x \cos y+ \sin x \sin y$
$\tan (x+y) = \dfrac{\tan x+\tan y}{1-\tan x \tan y}\\ \tan (x-y) = \dfrac{\tan x-\tan y}{1+\tan x \tan y}$
Work Step by Step
We have the following formulas:
$\sin (x+y)= \sin x \cos y+\cos x \sin y\\ \cos (x+y)= \cos x \cos y- \sin x \sin y\\ \sin (x-y)= \sin x \cos y -\cos x \sin y \\ \cos (x-y)= \cos x \cos y+ \sin x \sin y$
$\tan (x+y) = \dfrac{\tan x+\tan y}{1-\tan x \tan y}\\ \tan (x-y) = \dfrac{\tan x-\tan y}{1+\tan x \tan y}$