Prove the identity. tan2xsec⁡x=sec⁡x−cos⁡x

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Sally Cresswell · Accepted Answer

The first thing that comes to my mind is I want everything in terms of sin's and cos's if I can, so let's multiply through by cosx so that we can at least get rid of that secx=1cosx term.
$(\frac{\tan^{2} x}{\sec x}) \cos x = 1 - \cos^{2} x$
Now that $1 - \cos^{2} x$ is popping out at me because I remember the Pythagorean identity $\sin^{2} x + \cos^{2} x = 1$
Using this we see that $1 - \cos^{2} x = \sin^{2} x$ and so what we're trying to prove again simplifies to $(\frac{\tan^{2} x}{\sec x}) \cos x = \sin^{2} x$
Now let's work on the LHS and see if we can show that it is equal to $\sin^{2} x$ . Firstly, let's unpack the definitions of tan and sec.
$(t a n^{2} x / s e c x) c o s x = (s i n^{2} x / c o s^{2} x / 1 / c o s x) * c o s x = s i n^{2} x$
which is exactly the modified RHS of the equation. Thus the identity is proven.

Prove the identity. tan^2x/secx=secx-cosx

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