Answer
$\dfrac{\cos^{2}x-\tan^{2}x}{\sin^{2}x}=\cot^{2}x-\sec^{2}x$
Work Step by Step
$\dfrac{\cos^{2}x-\tan^{2}x}{\sin^{2}x}=\cot^{2}x-\sec^{2}x$
Rewrite the left side of the equation as $\dfrac{\cos^{2}x}{\sin^{2}x}-\dfrac{\tan^{2}x}{\sin^{2}x}$:
$\dfrac{\cos^{2}x}{\sin^{2}x}-\dfrac{\tan^{2}x}{\sin^{2}x}=\cot^{2}x-\sec^{2}x$
Substitute $\tan^{2}x$ by $\dfrac{\sin^{2}x}{\cos^{2}x}$:
$\dfrac{\cos^{2}x}{\sin^{2}x}-\dfrac{\Big(\dfrac{\sin^{2}x}{\cos^{2}x}\Big)}{\sin^{2}x}=\cot^{2}x-\sec^{2}x$
Evaluate $\dfrac{\Big(\dfrac{\sin^{2}x}{\cos^{2}x}\Big)}{\sin^{2}x}$:
$\dfrac{\cos^{2}x}{\sin^{2}x}-\dfrac{1}{\cos^{2}x}=\cot^{2}x-\sec^{2}x$
Since $\dfrac{\cos^{2}x}{\sin^{2}x}=\cot^{2}x$ and $\dfrac{1}{\cos^{2}x}=\sec^{2}x$, the identity is proved:
$\cot^{2}x-\sec^{2}x=\cot^{2}x-\sec^{2}x$
