Proving $\tan^2(x)-\sin^2(x) = \tan^2(x)\sin^2(x)$ (and trig identities in general)

Question

The problem is $$\tan^2(x)-\sin^2(x) = \tan^2(x)\sin^2(x)$$

I asked a question like this previously, and understood that one once someone helped me, but now I am back to not understanding with this problem. Could someone show me the steps and help me understand?

Also, I know it's a part of memorizing the trig identities and becoming familiar with them, but if anyone has any tips or something that can help me that would be great too.

Answer 1

General hint: convert everything to sines and cosines, and then there are only a couple of identities to mess with. In this case $$ \tan^2 x - \sin^2 x = \sin^2 x( \frac{1}{\cos^2 x} - 1) $$ Multiplying top and bottom by $\cos^2 x$, you get $$ \sin^2 x( \frac{1}{\cos^2 x} - 1) \frac{\cos^2 x}{\cos^2 x} = \sin^2 x \frac{1}{\cos^2 x}(1 - \cos^2 x)= \tan^2 x \sin^2 x. $$

Answer 2

$x^2-y^2$ is always $(x-y)(x+y)$, no matter how complicated $x$ and $y$ may be. So use that, then see what happens when you re-write $\tan x$ in terms of $\sin x$ and $\cos x$. Factor out what you can. That should get you on the way.