Proving $\frac{\tan x+\sec x-1}{\tan x-\sec x+1}=\frac{1+\sin x}{\cos x}$

Question

How do I myself start off this question:

Show that: $$\frac{\tan x+\sec x-1}{\tan x-\sec x+1}=\frac{1+\sin x}{\cos x}$$

I have tried to express the LHS in terms of $\sin x$ and $\cos x$, simplified the resulting expression, squared numerator and denominator and simplified again, by this time I was lost in the forest…

Answer 1

To solve question covert 1 in numerator to $\sec^2 x-\tan^2 x$.

Answer 2

I think, it is enough to multiply numerator and denominator on the left by $\cos {x}$.

Then you can cross-multiply sides.

$$\frac{\sin{x}+1-\cos{x}}{\sin{x}-1+\cos{x}} = \frac{1+\sin{x}}{\cos{x}}$$

Recall, $$\cos^2{x} = 1 - \sin^2{x}$$