Question

In problems 19-98, establish each identity.
$\frac{{\sin }^2\left(\theta \right)-\tan \left(\theta \right)}{{\cos }^2\left(\theta \right)-\cot \left(\theta \right)}={\tan }^2\left(\theta \right)$

Short answer

We have proved the identity$\frac{{\sin }^2\left(\theta \right)-\tan \left(\theta \right)}{{\cos }^2\left(\theta \right)-\cot \left(\theta \right)}={\tan }^2\left(\theta \right)$.

Step-by-step solution

Step 1. Given information.

We have:

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

Step 2. Establish the identity.

Take the left hand side:

$$\begin{gathered} \frac{{\sin }^2\left(\theta \right)-\tan \left(\theta \right)}{{\cos }^2\left(\theta \right)-\cot \left(\theta \right)} \\ =\frac{{\sin }^2\left(\theta \right)-{\displaystyle \frac{\sin \left(\theta \right)}{\cos \left(\theta \right)}}}{{\cos }^2\left(\theta \right)-{\displaystyle \frac{\cos \left(\theta \right)}{\sin \left(\theta \right)}}} \\ =\frac{{\displaystyle \frac{{\sin }^2\left(\theta \right)\cos \left(\theta \right)-\sin \left(\theta \right)}{\cos \left(\theta \right)}}}{{\displaystyle \frac{{\cos }^2\left(\theta \right)\sin \left(\theta \right)-\cos \left(\theta \right)}{\sin \left(\theta \right)}}} \\ =\frac{\sin \left(\theta \right)\bcancel{\left[\sin \left(\theta \right)\cos \left(\theta \right)-1\right]}}{\cos \left(\theta \right)}\times \frac{\sin \left(\theta \right)}{\cos \left(\theta \right)\bcancel{\left[\sin \left(\theta \right)\cos \left(\theta \right)-1\right]}} \\ =\frac{{\sin }^2\left(\theta \right)}{{\cos }^2\left(\theta \right)} \\ ={\tan }^2\left(\theta \right) \end{gathered}$$

Hence, proved.