# Simplify? 1/2 (tantheta-cottheta)/(tantheta+cottheta+1)

##### 1 Answer

I ended up with $- \cos \frac{2 x}{2 + \sin \left(2 x\right)}$

#### Explanation:

$\frac{1}{2} \frac{\tan \theta - \cot \theta}{\tan \theta + \cot \theta + 1}$

(sintheta/costheta-costheta/sintheta)/(2(sintheta/costheta+costheta/sintheta+1)

$\frac{\frac{{\sin}^{2} \theta - {\cos}^{2} \theta}{\sin \theta \cos \theta}}{\frac{2 {\sin}^{2} \theta + 2 {\cos}^{2} \theta + 2 \sin \theta \cos \theta}{\sin \theta \cos \theta}}$

$\left(\frac{{\sin}^{2} \theta - {\cos}^{2} \theta}{\sin \theta \cos \theta}\right) \times \left(\frac{\sin \theta \cos \theta}{2 {\sin}^{2} \theta + 2 {\cos}^{2} \theta + 2 \sin \theta \cos \theta}\right)$

$\frac{{\sin}^{2} \theta - {\cos}^{2} \theta}{2 {\sin}^{2} \theta + 2 {\cos}^{2} \theta + 2 \sin \theta \cos \theta}$

Recall that #sin^2x+cos^2x=1

$\frac{{\sin}^{2} \theta - {\cos}^{2} \theta}{2 + 2 \sin \theta \cos \theta}$

Recall that $\cos \left(2 x\right) = {\cos}^{2} x - {\sin}^{2} x \implies - \cos \left(2 x\right) = {\sin}^{2} x - {\cos}^{2} x$

$- \cos \frac{2 x}{2 + \sin \left(2 x\right)}$