## Calculus: Early Transcendentals 8th Edition

$tan^{2}\alpha-sin^{2}\alpha=tan^{2}\alpha$ $sin^{2}\alpha$
We need to prove the identity $tan^{2}\alpha-sin^{2}\alpha=tan^{2}\alpha$ $sin^{2}\alpha$ $tan^{2}\alpha-sin^{2}\alpha= \frac{sin^{2}\alpha}{cos^{2}\alpha}-sin^{2}\alpha$ $tan^{2}\alpha-sin^{2}\alpha=\frac{sin^{2}\alpha-sin^{2}\alpha cos^{2}\alpha}{cos^{2}\alpha}$ Thus, $tan^{2}\alpha-sin^{2}\alpha=\frac{1-cos^{2}}{cos^{2}\alpha}\times sin^{2}\alpha$ ${1-cos^{2}\alpha}=sin^{2}\alpha$ Now, $tan^{2}\alpha-sin^{2}\alpha=\frac{sin^{2}\alpha}{cos^{2}\alpha}\times sin^{2}\alpha$ Hence, $tan^{2}\alpha-sin^{2}\alpha=tan^{2}\alpha$ $sin^{2}\alpha$