Question

(\tan x + \cot x)^2 = \sec^2 x + \csc^2 x

Explanation:

Step1: Expand the left - hand side

We know that \((a + b)^2=a^{2}+2ab + b^{2}\). For \((\tan x+\cot x)^{2}\), let \(a = \tan x\) and \(b=\cot x\). Then \((\tan x+\cot x)^{2}=\tan^{2}x + 2\tan x\cot x+\cot^{2}x\).

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

So the left - hand side becomes \(\tan^{2}x+2+\cot^{2}x\).

Step2: Use trigonometric identities

We know the following trigonometric identities: \(\tan^{2}x+ 1=\sec^{2}x\), which implies \(\tan^{2}x=\sec^{2}x - 1\); and \(\cot^{2}x + 1=\csc^{2}x\), which implies \(\cot^{2}x=\csc^{2}x-1\).

Substitute \(\tan^{2}x=\sec^{2}x - 1\) and \(\cot^{2}x=\csc^{2}x - 1\) into \(\tan^{2}x+2+\cot^{2}x\):

$$
LATEXBLOCK0
$$

Which is equal to the right - hand side.

Answer:

The identity \((\tan x+\cot x)^{2}=\sec^{2}x+\csc^{2}x\) is proven.