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

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.

• 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 Math.StackExchange.com... – 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

## 3 Answers

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

• A perfectly explained solution. – James Jul 16 '12 at 22:26

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 $\displaystyle \frac{\cos^2x}{\cos^2x}$.

• 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
• 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

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

• See here for reference. – draks ... Jul 16 '12 at 22:19