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
 A: *

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

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

A: Note first that IF $\,\sin(x)= 0,\,$ then clearly also $\tan(x) = 0$ in which case the equation evaluates to $\,0 = 0\,$ which certainly is true. 
Now, in the case that $\,\sin x\neq 0\,$ (so also $\tan x\neq 0$), we can divide both sides of the equation  $$\tan^2 x - \sin^2 x = \sin^2 x\tan^2 x$$ by $\sin^2 x\tan^2 x,\,$ giving us $$\csc^2 x - \cot^2x = 1$$
But $\csc^2 x = 1 + \cot^2 x,\,$ so that gives us $$1 + \cot^2 x - \cot^2 x = 1,$$ as desired.
A: For $\#1$ $$\sin^2x+\sin^2x\cdot\tan^2x=\sin^2x(1+\tan^2x)=\frac{\sin^2x}{\cos^2x}$$
$$\implies \sin^2x+\sin^2x\cdot\tan^2x=\tan^2x$$
For $\#2$  $$\csc x=\frac1{\sin x},\sec x=\frac1{\cos x}$$
A: I just wanted to point out that when doing proofs like this that your argument should lead from the initial form (e.g. $\tan ^2 x - \sin^2 x$) to the final form ($\sin^2x \tan^2 x$). That should be your final answer, but sometimes it is just easier to prove that the LHS and RHS are equal.
For example, my doodles may look like this
$$\tan ^2 x - \sin^2 x = \sin^2x \tan^2 x$$
$$\tan ^2 x  = \sin^2x \tan^2 x + \sin^2 x$$
$$\tan ^2 x  = \sin^2x(\tan^2 x + 1)$$
$$\tan ^2 x  = \sin^2x(\sec^2 x)$$
$$\tan ^2 x  = \sin^2x\bigg(\frac{1}{\cos^2 x}\bigg)$$
$$\tan ^2 x  = \tan ^2 x$$
Now, I'm going to use my doodles to present my real argument (the one I hand into my teacher).
Recall that I ended up $\tan^2 x$, but started with $\tan ^2 x - \sin^2 x$, so I'm just going to take $\sin^2 x$ along for the ride for a bit, and work backwards from my doodles.
$$\tan ^2 x - \sin^2 x = \frac{\sin^2 x}{\cos^2 x}-\sin^2 x = \sin^2 x \sec^2 x-\sin^2 x = \sin^2 x(\sec^2 x-1) = \sin^2 x \tan^2 x $$
Try doing the same for the other problem.
By the way, this is how I remember the Pythagorean identities.

It's unfortunate that Geometry has been downplayed so much in our high schools in recent years.
