Question

(sinx-cosx)(sinx+cosx)/sin^2x-1/sec^2x

Answer 1

And what the heck are we supposed to do with this nonsense?

Answer 2

Simplify.

Answer 3

\(
\Large \frac{(\sin x - \cos x)(\sin x + \cos x)}{\left ( \sin^2 x - \frac{1}{\sec^2 x}\right)} = \Large \frac{(\sin x - \cos x)(\sin x + \cos x)}{\left ( \sin^2 x - \cos^2 x\right)} = \\ \Large
\Large \frac{(\sin x - \cos x)(\sin x + \cos x)}{(\sin x + \cos x)(\sin x - \cos x)} =
1
\)

Answer 4

Since \(a^2 - b^2 = (a+b)(a-b)\)

Answer 5

Ok