# Math Help - Proving identity

1. ## Proving identity

Prove that $1 - \frac{\cos{x}}{\sec{x}} = \sin^2x$
So I try to prove this by starting from the left hand side of the equation:
$LHS = 1 - \frac{\cos{x}}{\sec{x}}$
$= \sin^2{x} + \cos^2{x} - \frac{\frac{\cos{x}}{1}}{\cos{x}}$
$= \sin^2{x} + \cos^2{x} - 1$
$= \sin^2{x} + \cos^2{x} - (\sin^2{x} + \cos^2{x})$
$= 0$

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

Any pointers? Thank you in advance!

2. This line:

$= \sin^2{x} + \cos^2{x} - \dfrac{\cos(x)}{\frac{1}{\cos(x)}}$

cannot become this line:

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

This:

$= \sin^2{x} + \cos^2{x} - \dfrac{\cos(x)}{\frac{1}{\cos(x)}}$

Becomes this:

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

3. OK, now I can easily solve the problem. Thank you!

4. No problem!