## Trigonometry (11th Edition) Clone

$$(\tan x+\cot x)^2-(\tan x-\cot x)^2=4$$
$$A=(\tan x+\cot x)^2-(\tan x-\cot x)^2$$ We can use the identity: $$a^2-b^2=(a-b)(a+b)$$ That means $$A=[(\tan x+\cot x)-(\tan x-\cot x)][(\tan x+\cot x)+(\tan x-\cot x)]$$ $$A=[\tan x+\cot x-\tan x+\cot x][\tan x+\cot x+\tan x-\cot x]$$ $$A=[2\cot x][2\tan x]$$ $$A=4\tan x\cot x$$ - As a Reciprocal Identity states: $$\cot x=\frac{1}{\tan x}$$ $$\cot x\tan x=1$$ Therefore, $$A=4\times1=4$$