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.

$$\frac{\left(\sec A-\tan A\right)\left(\sec A+\tan A\right)}{\csc A-\cot A}=\cot A+\csc A$$ $$\left(\sec A-\tan A\right)\left(\sec A+\tan A\right)=\left(\csc A-\cot A\right)\left(\csc A+\cot A\right)$$ $$\sec^2A-\tan^2A=\csc^2A-\cot^2A$$ $$\frac{1}{\cos^2A}-\frac{\sin^2A}{\cos^2A}=\frac{1}{\sin^2A}-\frac{\cos^2A}{\sin^2A}$$ $$\frac{1-\sin^2A}{\cos^2A}=\frac{1-\cos^2A}{\sin^2A}$$ $$\frac{\cos^2A}{\cos^2A}=\frac{\sin^2A}{\sin^2A}$$ $$1=1$$

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.

in the third step rationalise by multiplying cosecA+cotA in numerator and denominator

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.