Answer
$tan^2~x-sin^2~x = (tan~x~sin~x)^2$
Work Step by Step
$tan^2~x-sin^2~x = \frac{sin^2~x}{cos^2~x}-sin^2~x$
$tan^2~x-sin^2~x = \frac{sin^2~x}{cos^2~x}-\frac{sin^2~x~cos^2~x}{cos^2~x}$
$tan^2~x-sin^2~x = \frac{sin^2~x-sin^2~x~cos^2~x}{cos^2~x}$
$tan^2~x-sin^2~x = (sin^2~x)~(\frac{1-cos^2~x}{cos^2~x})$
$tan^2~x-sin^2~x = (sin^2~x)~(\frac{sin^2~x}{cos^2~x})$
$tan^2~x-sin^2~x = (sin^2~x)~(tan^2~x)$
$tan^2~x-sin^2~x = (tan~x~sin~x)^2$
