Answer
$\dfrac{\cos^{2}t+\tan^{2}t-1}{\sin^{2}t}=\tan^{2}t$
Work Step by Step
$\dfrac{\cos^{2}t+\tan^{2}t-1}{\sin^{2}t}=\tan^{2}t$
On the left side, replace $1$ by $\sin^{2}t+\cos^{2}t$ and simplify the numerator:
$\dfrac{\cos^{2}t+\tan^{2}t-\sin^{2}t-\cos^{2}t}{\sin^{2}t}=\tan^{2}t$
$\dfrac{\tan^{2}t-\sin^{2}t}{\sin^{2}t}=\tan^{2}t$
Divide both terms in the numerator by $\sin^{2}t$:
$\dfrac{\tan^{2}t}{\sin^{2}t}-1=\tan^{2}t$
Replace $\tan^{2}t$ by $\dfrac{\sin^{2}t}{\cos^{2}t}$ and simplify:
$\dfrac{\sin^{2}t}{\sin^{2}t\cos^{2}t}-1=\tan^{2}t$
$\dfrac{1}{\cos^{2}t}-1=\tan^{2}t$
Replace $\dfrac{1}{\cos^{2}t}$ by $\sec^{2}t$:
$\sec^{2}t-1=\tan^{2}t$
Since $\sec^{2}t-1=\tan^{2}t$, the identity is proved:
$\tan^{2}t=\tan^{2}t$
