Proving a trigonometric identity (Homework) $$\frac{(\sec A - \tan A)(\sec A + \tan A)} {\csc A-\cot A} \equiv \cot A + \csc A $$

So I started by using DOTS (Difference of two squares) on the numerator on the left hand side. This gave me:

$$\frac{\sec^2x - \tan^2x}{\csc A - \cot A}$$

Then using the trigonometric identity: $\tan^2x + 1 = \sec^2x$
I solved to:

$$\frac{\sec^2x - (\sec^2x - 1)}{\csc A - \cot A}$$

And in turn:

$$\frac {1}{\csc A - \cot A}$$

=> $$\frac{1}{(\frac{1}{\sin A})} - \frac{1}{(\frac{1}{\tan A})}$$

=> $$\sin A - \tan A$$

From here I couldn't think of a way to get closer to $\cot A + \csc A$
Am I on the right track? If not, could someone point me in the right direction? Thanks.
 A: Just after "And in turn", you did something wrong: you wrote $\frac{1}{a - b} = \frac{1}{a} - \frac{1}{b}$, which is not generally true. 
But you're really close. Try substituting $\sin/\cos$ for $\tan$, and in fact, write all the other things in terms of sines and cosines, just so you have fewer things to worry about. 
A: $$\frac{\left(\sec A-\tan A\right)\left(\sec A+\tan A\right)}{\csc A-\cot A}=\cot A+\csc A$$
$$\left(\sec A-\tan A\right)\left(\sec A+\tan A\right)=\left(\csc A-\cot A\right)\left(\csc A+\cot A\right)$$
$$\sec^2A-\tan^2A=\csc^2A-\cot^2A$$
$$\frac{1}{\cos^2A}-\frac{\sin^2A}{\cos^2A}=\frac{1}{\sin^2A}-\frac{\cos^2A}{\sin^2A}$$
$$\frac{1-\sin^2A}{\cos^2A}=\frac{1-\cos^2A}{\sin^2A}$$
$$\frac{\cos^2A}{\cos^2A}=\frac{\sin^2A}{\sin^2A}$$
$$1=1$$
A: in the third step rationalise by multiplying cosecA+cotA in numerator and denominator
A: I always find these identities more straightforward if I get them in terms of $\sin A$ and $\cos A$
So $\sec A\pm \tan A=\cfrac 1{\cos A}\cdot(1\pm \sin A)$ so the numerator becomes $$\frac 1{\cos^2 A}(1-\sin^2 A)=1$$
Also $(\csc A+\cot A)(\csc A - \cot A)=\cfrac 1{\sin^2 A} (1-\cos^2 A)=1$ so that $$(\sec A+\tan A)(\sec A-\tan A)=(\csc A+\cot A)(\csc A - \cot A)$$
And aside from a technicality about not dividing by zero, this is what you need.
