# Prove cot^2theta/(sintheta+costheta)=(cos^2thetasintheta-cos^3theta)/(2sin^4theta-sin^2theta)?

See below:

#### Explanation:

To help follow what's going on I'll put in red text things that have changed from the line before.

${\cot}^{2} \frac{\theta}{\sin \theta + \cos \theta} = \frac{{\cos}^{2} \theta \sin \theta - {\cos}^{3} \theta}{2 {\sin}^{4} \theta - {\sin}^{2} \theta}$

I'll first use ${\cot}^{2} \theta = {\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta$ on the left side and factor the numerator and denominator on the right side. We can also see a $\sin \theta - \cos \theta$ term setting up in the right numerator and so I'll multiply through on the left using that term:

((color(red)(cos^2theta/sin^2theta))/(sintheta+costheta))color(red)(((sintheta-costheta)/(sintheta-costheta)))=color(red)((cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1))

cos^2theta/((sin^2theta)(sintheta+costheta))((sintheta-costheta)/(sintheta-costheta))=(cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1)

(cos^2thetacolor(red)((sintheta-costheta)))/((color(red)(sin^2theta))(sintheta+costheta)color(red)((sintheta-costheta)))=(cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1)

(cos^2theta(sintheta-costheta))/((sin^2theta)color(red)((sin^2theta-cos^2theta)))=(cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1)

We can now use ${\sin}^{2} \theta + {\cos}^{2} \theta = 1 \implies {\cos}^{2} \theta = 1 - {\sin}^{2} \theta$

(cos^2theta(sintheta-costheta))/((sin^2theta)(sin^2theta-color(red)((1-sin^2theta))))=(cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1)

(cos^2theta(sintheta-costheta))/((sin^2theta)(sin^2thetacolor(red)(-1+sin^2theta)))=(cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1)

(cos^2theta(sintheta-costheta))/((sin^2theta)color(red)((2sin^2theta-1)))=(cos^2theta(sintheta-costheta))/(sin^2theta(2sin^2theta-1)

Jan 14, 2017

$L H S = \frac{{\cos}^{2} \theta \sin \theta - {\cos}^{3} \theta}{2 {\sin}^{4} \theta - {\sin}^{2} \theta}$

$= \frac{{\cos}^{2} \theta \left(\sin \theta - \cos \theta\right)}{{\sin}^{2} \theta \left(2 {\sin}^{2} \theta - 1\right)}$
$= \frac{{\cot}^{2} \theta \left(\sin \theta - \cos \theta\right)}{{\sin}^{2} \theta - 1 + {\sin}^{2} \theta}$

$= \frac{{\cot}^{2} \theta \left(\sin \theta - \cos \theta\right)}{{\sin}^{2} \theta - \left(1 - {\sin}^{2} \theta\right)}$

$= \frac{{\cot}^{2} \theta \left(\sin \theta - \cos \theta\right)}{{\sin}^{2} \theta - {\cos}^{2} \theta}$

=(cot^2theta(sintheta-costheta))/((sintheta-costheta)(sintheta+costheta)

$= {\cot}^{2} \frac{\theta}{\sin \theta + \cos \theta} = L H S$

Proved