<s>[INST] Proving some trig identities. Stuck on two questions.
First one: $\tan^2x - \sin^2x = (\sin^2 x)(\tan^2 x)$
Tried solving that with the right side but wasn't able to.
Second one: $\csc x / \sec x = \cot x$
I tried solving this one with the left side doing,
L.S = $\csc x / \sec x $
L.S = $(1 / \sin x) / (1 / \cos x) $
L.S = $\sin x / \cos x\quad$    [(1's cancel out]
L.S = $\tan x$
Don't know what I'm doing wrong here.
For reference, here is the trig identity worksheet we're supposed to use.
http://imgur.com/TXxaYUQ
 [/INST] *

*$$
\begin{align}
\tan^2x - \sin^2x&=\frac{\sin^2x}{\cos^2x}-\sin^2x\quad;\ \text{where}\ \tan x=\frac{\sin x}{\cos x}\\
&=\sin^2x\left(\frac{1}{\cos^2x}-1\right)\\
&=\sin^2x\left(\frac{1}{\cos^2x}-\frac{\cos^2x}{\cos^2x}\right)\\
&=\frac{\sin^2x}{\cos^2x}\left(1-\cos^2x\right)\\
&=\color{blue}{\tan^2x \cdot \sin^2x}\quad;\ \text{where}\ \sin^2x + \cos^2x=1.
\end{align}
$$
$$$$

*$$
\begin{align}
\frac{\csc x}{\sec x}&=\dfrac{\dfrac{1}{\sin x}}{\dfrac{1}{\cos x}}\\
&=\frac{1}{\color{red}{\sin x}}\times\frac{\color{red}{\cos x}}{1}\\
&=\frac{\cos x}{\sin x}\\
&=\color{blue}{\cot x}.
\end{align}
$$

</s>