Question

factor the trigonometric expression.
$ ( cot x + \tan x ) ^ { 2 } - ( cot x - \tan x ) ^ { 2 } $
$ ( cot x + \tan x ) ^ { 2 } - ( cot x - \tan x ) ^ { 2 } = square $ (simplify your answer.)

Explanation:

Step1: Use the difference of squares formula \(a^{2}-b^{2}=(a + b)(a - b)\)

Let \(a=\cot x+\tan x\) and \(b=\cot x-\tan x\). Then \((\cot x+\tan x)^{2}-(\cot x - \tan x)^{2}=[(\cot x+\tan x)+(\cot x-\tan x)][(\cot x+\tan x)-(\cot x-\tan x)]\)

Step2: Simplify the expressions inside the brackets

First bracket: \((\cot x+\tan x)+(\cot x-\tan x)=\cot x+\tan x+\cot x-\tan x = 2\cot x\)
Second bracket: \((\cot x+\tan x)-(\cot x-\tan x)=\cot x+\tan x-\cot x+\tan x=2\tan x\)

Step3: Multiply the two simplified expressions

\(2\cot x\times2\tan x\). Since \(\cot x=\frac{\cos x}{\sin x}\) and \(\tan x=\frac{\sin x}{\cos x}\), then \(\cot x\times\tan x = 1\)

Answer:

\(4\)