Answer
$tan^{2}\alpha-sin^{2}\alpha=tan^{2}\alpha$ $sin^{2}\alpha$
Work Step by Step
Need to prove the identity $tan^{2}\alpha-sin^{2}\alpha=tan^{2}\alpha$ $sin^{2}\alpha$
In order to prove this, take left side isolate.
$tan^{2}\alpha-sin^{2}\alpha= \frac{sin^{2}\alpha}{cos^{2}\alpha}-sin^{2}\alpha$
$tan^{2}\alpha-sin^{2}\alpha=\frac{sin^{2}\alpha-sin^{2}\alpha cos^{2}\alpha}{cos^{2}\alpha}$
Thus,
$tan^{2}\alpha-sin^{2}\alpha=\frac{1-cos^{2}}{cos^{2}\alpha}\times sin^{2}\alpha$
Since, ${1-cos^{2}\alpha}=sin^{2}\alpha$,
Now, $tan^{2}\alpha-sin^{2}\alpha=\frac{sin^{2}\alpha}{cos^{2}\alpha}\times sin^{2}\alpha$
Hence,$tan^{2}\alpha-sin^{2}\alpha=tan^{2}\alpha$ $sin^{2}\alpha$
