# 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?

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Multiply the numerator and denominator of the first fraction in the LHS by $\cos x$ (highlighted in blue), so that we have $$\color{blue}{\frac{\tan x}{\sec x - 1}}+\frac{1-\cos x}{\sin x}=\color{blue}{\frac{\sin x}{1-\cos x}}+\frac{1-\cos x}{\sin x}\\=\frac{\sin^2 x+1-2\cos x+\cos^2x}{\sin x(1-\cos x)}$$ Use the identity that $\sin^2x+\cos^2x=1$. Can you take it from here?

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Thanks! Got it now. – Aspiring Mathlete Jun 10 '14 at 19:22

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

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HINT:

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

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