Prove the following Trig Identity with reciprocals Prove that:
$$\frac{\tan x}{\sec x-1}+\frac{1-\cos x}{\sin x}=2\csc x$$
Help please! I tried so many things but couldn't get the LHS = RHS. A hint please?

Hint:
$$
(1 - \cos x)(1 + \cos x) = 1 - \cos^2 x = \sin^2 x\\
(\sec x - 1)(\sec x + 1) = \sec^2 x - 1 = \tan^2 x
$$
Further
$$
\frac{\sec x + 1}{\tan x} = \frac{1+\cos x}{\sin x} = \frac{(1 + \cos x)^2}{(1 + \cos x) \sin x}\\
\frac{\sin x}{1 + \cos x} = \frac{\sin^2 x}{(1 + \cos x) \sin x}
$$
