Answer
$\tan^{2}u-\sin^{2}u=\tan^{2}u\sin^{2}u$
Work Step by Step
$\tan^{2}u-\sin^{2}u=\tan^{2}u\sin^{2}u$
On the left side of the equation, replace $\tan^{2}u$ by $\dfrac{\sin^{2}u}{\cos^{2}u}$:
$\dfrac{\sin^{2}u}{\cos^{2}u}-\sin^{2}u=\tan^{2}u\sin^{2}u$
Evaluate the subtraction on the left side:
$\dfrac{\sin^{2}u-\sin^{2}u\cos^{2}u}{\cos^{2}u}=\tan^{2}u\sin^{2}u$
Take out common factor $\sin^{2}u$ from the numerator:
$\dfrac{\sin^{2}u(1-\cos^{2}u)}{\cos^{2}u}=\tan^{2}u\sin^{2}u$
Rewrite the left side of the equation as $\dfrac{\sin^{2}u}{\cos^{2}u}\cdot(1-\cos^{2}u)$:
$\dfrac{\sin^{2}u}{\cos^{2}u}\cdot(1-\cos^{2}u)=\tan^{2}u\sin^{2}u$
Since $\dfrac{\sin^{2}u}{\cos^{2}u}=\tan^{2}u$ and $1-\cos^{2}u=\sin^{2}u$, the identity is proved
$\tan^{2}u\sin^{2}u=\tan^{2}u\sin^{2}u$
