# How do you prove csc^2theta-cot^2theta=cotthetatantheta?

##### 2 Answers
Dec 27, 2016

Use the identities $\csc \beta = \frac{1}{\sin} \beta$, $\cot \beta = \cos \frac{\beta}{\sin} \beta$ and $\tan \beta = \sin \frac{\beta}{\cos} \beta$.

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

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

Use the identity ${\sin}^{2} \beta + {\cos}^{2} \beta = 1$.

${\sin}^{2} \frac{\theta}{\sin} ^ 2 \theta = 1$

$1 = 1$

$L H S = R H S$

Hopefully this helps!

Dec 27, 2016

${\csc}^{2} \theta - {\cot}^{2} \theta = \cot \theta \tan \theta$

Use the Pythagorean identity $1 + {\cot}^{2} \theta = {\csc}^{2} \theta$

$1 + {\cot}^{2} \theta - {\cot}^{2} \theta = \cot \theta \tan \theta$

$1 = \cot \theta \tan \theta$

$\tan \frac{\theta}{\tan} \theta = \cot \theta \tan \theta$

$\tan \theta \cdot \frac{1}{\tan} \theta = \cot \theta \tan \theta$

Use the identity $\cot \theta = \frac{1}{\tan} \theta$

$\tan \theta \cdot \cot \theta = \cot \theta \tan \theta$

$\cot \theta \tan \theta = \cot \theta \tan \theta$