Question

Show that ${\cos }^4\alpha +2{\cos }^2\alpha (1-\frac{1}{{\sec }^2\alpha })=1-{\sin }^4\alpha$

Answer 1

LHS$={\cos }^4\alpha +2{\cos }^2\alpha (1-\frac{1}{{\sec }^2\alpha })$

$={\cos }^4\alpha +2{\cos }^2\alpha (1-{\cos }^2\alpha )$

$={\cos }^4\alpha +2{\cos }^2\alpha -2{\cos }^4\alpha$

$=2{\cos }^2\alpha -{\cos }^4\alpha$

$={\cos }^2\alpha (2-{\cos }^2\alpha )$

$=(1-{\sin }^2\alpha )(2-1+{\sin }^2\alpha )$

$=(1-{\sin }^2\alpha )(1+{\sin }^2\alpha )$

$=1-{\sin }^4\alpha$

$=RHS$

Hence proved.