Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

How do I verify the following identity:

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

I have done simpler problems but got stuck with this one. Please help.


share|cite|improve this question
are you sure your brackets are in the right place? – oks Mar 7 '13 at 18:19
Multiple across by $\sin x$ and use a basic relationship involving $\sin^2 x + \cos^2 x$. – copper.hat Mar 7 '13 at 18:22
What you've written isn't an identity, it's an equation. An equation asks "For which values of $x$ are the left- and right-hand sides equal." For an identity, you need to use $\equiv$. For example $\cos^2x+\sin^2 \equiv 1$. This means that the left- and right-hand sides are identical. So writing $\cos^2x + \sin^2x \equiv 1$ is the same as saying that $\sin^x+\cos^2=1$ for all values of $x$. – Fly by Night Mar 7 '13 at 18:25
@FlybyNight: An identity (according to every trig book I've ever seen) is is an equation that holds whenever both sides of the equation are defined. This is an identity in that sense. – Cameron Buie Mar 7 '13 at 20:53
@CameronBuie I agree 100%. If $\sin^2x + \cos^2x = 1$ for all $x$ then $\sin^2x + \cos^2x \equiv 1$. (Which is of course true.) However, an equation, in general, is not an identity. – Fly by Night Mar 7 '13 at 21:07

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

Here I used the identity: $\sin^2x+\cos^2x=1$

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.