Answer
$$(\tan x+\cot x)^2-(\tan x-\cot x)^2=4$$
Work Step by Step
$$A=(\tan x+\cot x)^2-(\tan x-\cot x)^2$$
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, $$A=4\times1=4$$