Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|cite|improve this question

3 Answers 3

up vote 3 down vote accepted

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?

share|cite|improve this answer
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 $$ \frac{\sec x + 1}{\tan x} = \frac{1+\cos x}{\sin x} = \frac{(1 + \cos x)^2}{(1 + \cos x) \sin x}\\ \frac{\sin x}{1 + \cos x} = \frac{\sin^2 x}{(1 + \cos x) \sin x} $$

share|cite|improve this answer


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

share|cite|improve this answer
@user108104, How about this? – lab bhattacharjee Jun 11 '14 at 15:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.