Answer
$\dfrac{1}{1-\sin^{2}y}=1+\tan^{2}y$
Work Step by Step
$\dfrac{1}{1-\sin^{2}y}=1+\tan^{2}y$
On the left side of the equation, substitute $1-\sin^{2}y$ with $\cos^{2}y$:
$\dfrac{1}{\cos^{2}y}=1+\tan^{2}y$
Substitute $\dfrac{1}{\cos^{2}y}$ with $sec^{2}y$:
$\sec^{2}y=1+\tan^{2}y$
Since $\sec^{2}y=1+\tan^{2}y$, the identity is proved:
$1+\tan^{2}y=1+\tan^{2}y$
