So I try to prove this by starting from the left hand side of the equation:Prove that $\displaystyle 1 - \frac{\cos{x}}{\sec{x}} = \sin^2x$

$\displaystyle LHS = 1 - \frac{\cos{x}}{\sec{x}}$

$\displaystyle = \sin^2{x} + \cos^2{x} - \frac{\frac{\cos{x}}{1}}{\cos{x}}$

$\displaystyle = \sin^2{x} + \cos^2{x} - 1$

$\displaystyle = \sin^2{x} + \cos^2{x} - (\sin^2{x} + \cos^2{x})$

$\displaystyle = 0$

But this is not $\displaystyle = \sin^2{x}$

Any pointers? Thank you in advance!