# Question 74e29

Mar 2, 2017

When in doubt, rewrite using sine and cosine. See below.

#### Explanation:

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

Factor out $\sin \theta$ on top and $\cos \theta$ on the bottom, to get a $\tan \theta$

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

For the remaining quotient, get a single fraction over another fraction. (No I'm not certain it will work, but I have to try something, so . . . )

$= \tan \left(\theta\right) \frac{\left(\frac{\sin \left(\theta\right) \cos \left(\theta\right) - 1}{\cos \left(\theta\right)}\right)}{\left(\frac{\sin \left(\theta\right) \cos \left(\theta\right) - 1}{\sin \left(\theta\right)}\right)}$

We now see that we have the same numerators, so when we invert the denominator and multiply, the ((sin(theta)cos(theta)-1)# will cancel and leave $\sin \frac{\theta}{\cos} \theta$

$= \tan \left(\theta\right) \left(\frac{\sin \left(\theta\right) \cos \left(\theta\right) - 1}{\cos \left(\theta\right)} \cdot \sin \frac{\theta}{\left(\sin \left(\theta\right) \cos \left(\theta\right) - 1\right)}\right)$

$= \tan \left(\theta\right) \sin \frac{\theta}{\cos} \left(\theta\right)$

$= {\tan}^{2} \left(\theta\right)$