Question

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

Trigonometric equation and identitie
Prove the identity. $$\displaystyle\frac{{{\tan}^{{2}}{x}}}{{\sec{{x}}}}={\sec{{x}}}-{\cos{{x}}}$$

2020-12-30

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.
$$\displaystyle{\left(\frac{{{\tan}^{{2}}{x}}}{{\sec{{x}}}}\right)}{\cos{{x}}}={1}-{{\cos}^{{2}}{x}}$$
Now that 1−cos^2x is popping out at me because I remember the Pythagorean identity $$\displaystyle{{\sin}^{{2}}{x}}+{{\cos}^{{2}}{x}}={1}$$
Using this we see that $$\displaystyle{1}−{{\cos}^{{2}}{x}}={{\sin}^{{2}}{x}}$$ and so what we're trying to prove again simplifies to $$\displaystyle{\left(\frac{{{\tan}^{{2}}{x}}}{{\sec{{x}}}}\right)}{\cos{{x}}}={{\sin}^{{2}}{x}}$$
Now let's work on the LHS and see if we can show that it is equal to $$\displaystyle{{\sin}^{{2}}{x}}$$. Firstly, let's unpack the definitions of tan and sec.
$$(tan^2x/secx)cosx=(sin^2x/cos^2x/1/cosx)*cosx =sin^2x$$
which is exactly the modified RHS of the equation. Thus the identity is proven.