# What is cot^4theta/(1-cos^4theta) in terms of non-exponential trigonometric functions?

Mar 14, 2018

$\frac{\cot \theta \cdot \cot \theta}{\sin \theta \cdot \sin \theta \cdot \sin \theta \cdot \sin \theta}$

#### Explanation:

${\cot}^{4} \theta = {\cos}^{4} \frac{\theta}{\sin} ^ 4 \theta$

$1 - {\cos}^{2} \theta = {\sin}^{2} \theta$

Here is where the mistake was made:
cot^4 theta / (1-cos^4 theta) = (cos^4 theta/sin^4 theta)/(sin^2 theta*cos^2 theta

$\frac{{\cos}^{4} \frac{\theta}{\sin} ^ 4 \theta}{{\sin}^{2} \theta \cdot {\cos}^{2} \theta} = \frac{{\cos}^{2} \frac{\theta}{\sin} ^ 4 \theta}{{\sin}^{2} \theta}$

$\frac{{\cos}^{2} \frac{\theta}{\sin} ^ 4 \theta}{{\sin}^{2} \frac{\theta}{1}}$ or ${\cos}^{2} \frac{\theta}{\sin} ^ 4 \theta \cdot \frac{1}{\sin} ^ 2 \theta$

Multiply across

${\cos}^{2} \frac{\theta}{\sin} ^ 6 \theta$ = ${\cot}^{2} \frac{\theta}{\sin} ^ 4 \theta$

Expand to get rid of the exponents

$\frac{\cot \theta \cdot \cot \theta}{\sin \theta \cdot \sin \theta \cdot \sin \theta \cdot \sin \theta}$

I hope this is the form you were looking for!

Mar 14, 2018

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

#### Explanation:

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

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