Pythagorean Trig Identity I'm about to teach basic Pythgorean trig identities and went through the textbook exercise but I'm stuck on one.  
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$$\sec^2 A = \frac{ \mathrm{cosec} A}{ \mathrm{cosec} A - \sin A}$$
Can someone help please?
Thanks
Rob
 A: $$\frac{\csc A}{\csc A- \sin A}$$
$$=\frac1{1-\sin^2A}(\text{multiplying the numerator & the denominator by }\sin A )$$
$$=\frac1{\cos^2A}=\sec^2A$$
A: In order to simplify this, I would first convert the $\csc A$ to $\frac{1}{\sin A}$. With this, I would get the following:
$$\begin{align}\frac{\frac{1}{\sin A}}{\frac{1}{\sin A}-\sin A}\end{align}$$
Now I can easily simplify from here by making the denominator between $\frac{1}{\sin A}$ and $\sin A$ the same. This would give me, for the denominator:
$$\begin{align}
\frac{1 - \sin^2 A}{\sin A}
\end{align}$$
I can convert $1-\sin^2 A$ to $\cos^2 A$ and I would get the following: 
$$\frac{\frac{1}{\sin A}}{\frac{\cos^2A}{\sin A}}$$
Now you can see that I can easily cancel out the $\sin A$ from the numerator and denominator and I would be left with $\frac{1}{\cos^2A}$ which is the same as $\mathrm{sec^2} A$
A: $$\frac{\csc A}{\csc A - \sin A} = \frac{\csc A}{\csc A - \sin A}\cdot\frac{\csc A}{\csc A} = \frac{\csc^2 A}{\csc^2 A -\sin A\csc A}$$
$$ = \frac{\csc^2 A}{\csc^2 A -1}=\frac{\csc^2 A}{\cot^2A}=\tan^2 A\csc^2A=\tan^2 A(\cot^2A+1)$$
$$=\tan^2 A \cot^2 A + \tan^2 A = 1 + \tan^2 A = \sec^2 A$$
