Question

P.T. ${tan}^{2}A-si{n}^{2}A={\sin }^{4}A{\sec }^{2}A$.

Answer 1

L.H.S$={\tan }^{2}A-{\sin }^{2}A$

$=\frac{{\sin }^{2}A}{{\cos }^{2}A}-{\sin }^{2}A$

$={\sin }^{2}A(\frac{1}{{\cos }^{2}A}-1)$

$={\sin }^{2}A\left(\frac{1-{\cos }^{2}A}{{\cos }^{2}A}\right)$

$={\sin }^{2}A\left(\frac{{\sin }^{2}A}{{\cos }^{2}A}\right)$

$={\sin }^{4}A{\sec }^{2}A=$R.H.S

Hence proved.