# Grade 12 trigonometric identities: $\frac{ \cos 2x}{1+\sin 2x}=\tan\left(\frac{\pi}{4}-x\right)$

$$\frac{ \cos 2x}{1+\sin 2x}=\tan\left(\frac{\pi}{4}-x\right)$$

Hi. I'm confused how to prove this trig identity for the left side. If someone could help, then that would be great! I tried using $\sin(2x)=2\sin x\cos x$.

• You don't "solve" an identity, you prove it. Start with the RHS. Use the formula for $\tan(x-y)$. Then, write everything in terms of sine and cosine and multiply both numerator and denominator by $(\cos x+\sin x)$. Recall the formulas for $\cos(2x)$ and $\sin(2x)$ and that $\sin^2 x+\cos^2 x=1$ and you're done. – learner Nov 28 '15 at 20:17

HINT: we have by the Addition formulas $$\frac{\cos(2x)}{1+\sin(2x)}=\frac{\cos(x)^2-\sin(x)^2}{1+2\sin(x)\cos(x)}=\frac{\cos(x)-\sin(x)}{\cos(x)+\sin(x)}$$
• Replace $1$ with $cos^2(x)+sin^2(x)$ from the Pythagorean theorem, so you now have $cos^2(x) + 2sin(x)cos(x) + sin^2(x)$ in the denominator. This looks like a quadratic that is a perfect square, or a binomial squared, that factors to $(cos(x) + sin(x))^2$. One of those factors then cancels with the numerator that factors to $(cos(x)+sin(x))(cos(x)-sin(x))$. – cr3 Nov 28 '15 at 21:54
L.H.S.=$\frac{\cos 2x}{1+\sin 2x}=\frac{\cos ^2 x-\sin ^2 x}{(\cos x+\sin x)^2}=\frac{\cos x-\sin x}{(\cos x+\sin x)}=\frac{1-\tan x}{1+ \tan x}=\tan(\frac{\pi}{4}-x)$=R.H.S.
$$\cos 2x=\frac{1-\tan^2 x}{1+\tan^2 x}, \quad \sin 2x=\frac{2\tan x}{1+\tan^2 x},\quad\tan\Bigl(\frac\pi4-x\Bigr)=\frac{1-\tan x}{1+\tan x}.$$