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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have the following trigonometric identity

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

I've been trying to verify it for almost 20 minutes but coming up with nothing

Thank you

share|cite|improve this question
up vote 4 down vote accepted

Observe that we need to eliminate $\tan x$

So, using $\tan x=\frac{\sin x}{\cos x},$

$$\cos x-\frac{\cos x}{1-\tan x}$$

$$=\cos x\left(1-\frac{1}{1-\frac{\sin x}{\cos x}}\right)$$

$$=\cos x\left(1-\frac{\cos x}{\cos x-\sin x}\right) (\text{ multiplying numerator & denominator by }\cos x)$$

$$=\cos x\left(1+\frac{\cos x}{\sin x-\cos x}\right)$$

$$=\cos x\left(\frac{\sin x}{\sin x-\cos x}\right)$$

share|cite|improve this answer
How is your final answer equal to the RHS? – user82124 Jun 12 '13 at 15:28
@user82124, how about multiplying $\cos x$ and $\left(\frac{\sin x}{\sin x-\cos x}\right)$ – lab bhattacharjee Jun 12 '13 at 15:29
That would give cos x sin x / sin x cos x^2 – user82124 Jun 12 '13 at 15:31
@user82124, where is this $\cos x\sin x/\sin x\cos x^2$ coming from? – lab bhattacharjee Jun 12 '13 at 15:32
From multiplying cos x and ( Sin x / Sin x - Cos X )? – user82124 Jun 12 '13 at 15:35

$$ \frac{\cos}{1-\tan x}\cdot\frac{\cos x}{\cos x} = \frac{\cos^2 x}{\cos x-\sin x} $$

$$ \cos x\cdot\frac{\cos x-\sin x}{\cos x-\sin x} - \frac{\cos^2 x}{\cos x-\sin x} = \frac{-\sin x\cos x}{\cos x-\sin x} $$

share|cite|improve this answer

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.