#### Answer

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

#### Work Step by Step

$$A=(\tan x+\cot x)^2$$
$$A=\tan^2x+2\tan x\cot x+\cot^2 x$$
$$A=\tan^2 x+\cot^2 x+2\tan x\cot x$$
- According to a Reciprocal Identity:
$$\cot x=\frac{1}{\tan x}$$
which means $$\cot x\tan x=1$$
Therefore, $A$ would be $$A=\tan^2 x+\cot^2 x+2\times1$$
$$A=\tan^2 x+\cot^2 x+2$$
$$A=(\tan^2 x+1)+(\cot^2 x+1)$$
- We have 2 following Pythagorean Identities:
$$\tan^2 x+1=\sec^2 x$$ and $$\cot^2 x+1=\csc^2 x$$
Apply them to $A$: $$A=\sec^2 x+\csc^2 x$$