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 would I simplify this difficult trigonometric identity:

$$\frac{\sin A \cos A}{\cos^2 A - \sin^2 A} = \frac{\tan A}{1-\tan^2 A}.$$

I am not exactly sure what to do.

I simplified the right side to

$$\frac{\frac{\sin A}{\cos A}}{1-\frac{\cos^2 A}{\sin^2 A}}$$

But how would I proceed.

share|cite|improve this question
Use $\sin 2A= 2\sin A \cos A$ and $ \cos 2A= \cos^2 A- \sin^2 A$. – draks ... Jul 16 '12 at 22:15
I will say all the post were helpful and I would give everyone recognition were I a registered user. – James Jul 16 '12 at 22:27
James, you can up/down vote on all posts (question, answers, comments) here (except your own ;-). Additionally you can choose among the given answer and accept the one that suits you most. Read the faq for more information. And last: Welcome to – draks ... Jul 16 '12 at 22:35
@draks, the ability to upvote requires 15 rep, which James had had for only about 2 minutes when he posted that comment. Quite possibly he hadn't noticed yet. Downvoting, on the other hand, requires 125 rep. – Henning Makholm Jul 16 '12 at 23:28
@HenningMakholm ah right, sorry, I forgot. – draks ... Jul 17 '12 at 7:02

Your error lies in how you simplified the right hand side (the denominator specifically). Try again! Turn $\tan x$ into $\sin x$ and $\cos x$ with $\displaystyle\tan x =\frac{\sin x}{\cos x}$.

Now, $$\frac{\tan{x}}{1-\tan^2{x}}=\frac{\left(\frac{\sin x}{\cos x}\right)}{1-\left(\frac{\sin x}{\cos x}\right)^2}$$ You can multiply an expression by 1 and not change the value, (Since $1\cdot a=a$). Now, the problem is which 1 do you multiply by?

You can achieve this by multiplying and distributing by $\displaystyle \frac{\cos^2x}{\cos^2x}$.

share|cite|improve this answer
I see I wrote it like that in my paper and now I have (sin/cos)/(1-sin^2/cos^2) but I am still stuck. – James Jul 16 '12 at 22:17
thanks for your response. – James Jul 16 '12 at 22:27
these identities are sometimes difficult but you can always simplify the strategy for solving them into two steps: 1) Turn everything into $\sin x$ and $\cos x$, and 2) Look for Pythagorean Identities (i.e. $\sin^2x+\cos^2x=1$, $\tan^2x+1=\sec^2x$, $\cot^2x+1=\csc^2x$) – JohnnyK Jul 16 '12 at 22:35
And the Pythagorean Identities part can be expanded to include $\cos^2x=1-\sin^2x=(1+\sin x)(1-\sin x)$. Anytime you see things like that try to find a Pythagorean identity – JohnnyK Jul 16 '12 at 22:37

$$RHS = \frac{\tan A}{1-\tan^2 A} = \frac{\frac{\sin A}{\cos A}}{1-\frac{\sin^2 A}{\cos^2 A}}$$ $$=\frac{\frac{\sin A}{\cos A}}{\frac{\cos^2 A - \sin^2 A}{\cos^2 A}}\cdot \frac{\cos^2 A}{\cos^2 A}$$ $$=\frac{\sin A \cos A}{\cos^2 A - \sin^2 A} = LHS$$

share|cite|improve this answer
A perfectly explained solution. – James Jul 16 '12 at 22:26

Use $\sin 2A= 2\sin A \cos A$ and $ \cos 2A= \cos^2 A- \sin^2 A$ to get $$ \frac{\sin A \cos A}{\cos^2 A- \sin^2 A}=\frac{\sin 2A}{2\cos 2A}=\frac12\tan{2A}, $$ which is equivalent to $$ \frac{ \tan A} {1 - \tan^2 A} . $$

share|cite|improve this answer
See here for reference. – draks ... Jul 16 '12 at 22:19
thanks for the help. – James Jul 16 '12 at 22:26

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.