Answer
$(\tan x+\cot x)^{2}=\csc^{2}x\sec^{2}x$
Work Step by Step
$(\tan x+\cot x)^{2}=\csc^{2}x\sec^{2}x$
Evaluate the power on the left side:
$\tan^{2}x+2\tan x\cot x+\cot^{2}x=\csc^{2}x\sec^{2}x$
Substitute $\tan^{2}x$ by $\sec^{2}x-1$ and $\cot^{2}x$ by $\csc^{2}x-1$. Also substitute $\tan x$ by $\dfrac{\sin x}{\cos x}$ and $\cot x$ by $\dfrac{\cos x}{\sin x}$, then simplify:
$\sec^{2}x-1+2\Big(\dfrac{\sin x}{\cos x}\Big)\Big(\dfrac{\cos x}{\sin x}\Big)+\csc^{2}x-1=\csc^{2}x\sec^{2}x$
$\sec^{2}x-2+2(1)+\csc^{2}x=\csc^{2}x\sec^{2}x$
$\sec^{2}x+\csc^{2}x=\csc^{2}x\sec^{2}x$
Substitute $\sec^{2}x$ by $\dfrac{1}{\cos^{2}x}$ and $\csc^{2}x$ by $\dfrac{1}{\sin^{2}x}$:
$\dfrac{1}{\cos^{2}x}+\dfrac{1}{\sin^{2}x}=\csc^{2}x\sec^{2}x$
Evaluate the sum on the left side:
$\dfrac{\sin^{2}x+\cos^{2}x}{\sin^{2}x\cos^{2}x}=\csc^{2}x\sec^{2}x$
Substitute $\sin^{2}x+\cos^{2}x$ by $1$:
$\dfrac{1}{\sin^{2}x\cos^{2}x}=\csc^{2}x\sec^{2}x$
Rewrite the left side as $\Big(\dfrac{1}{\sin^{2}x}\Big)\Big(\dfrac{1}{\cos^{2}x}\Big)$:
$\Big(\dfrac{1}{\sin^{2}x}\Big)\Big(\dfrac{1}{\cos^{2}x}\Big)=\csc^{2}x\sec^{2}x$
Since $\dfrac{1}{\sin^{2}x}=\csc^{2}x$ and $\dfrac{1}{\cos^{2}x}=\sec^{2}x$, the identity is proved.
$\csc^{2}x\sec^{2}x=\csc^{2}x\sec^{2}x$
