How do you express sin^2 theta - sec^2 thetacottheta + tan^2 theta  in terms of cos theta ?

Jun 30, 2018

$\frac{1 - {\cos}^{4} \left(\theta\right)}{\cos} ^ 2 \left(\theta\right) - \frac{1}{\cos \left(\theta\right) \sin \left(\theta\right)}$ where

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

Explanation:

Using

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

${\tan}^{2} \left(\theta\right) = {\sin}^{2} \frac{\theta}{\cos} ^ 2 \left(\theta\right) = \frac{1 - {\cos}^{2} \left(\theta\right)}{\cos} ^ 2 \left(\theta\right)$
putting Things together

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

this is equal to

$\frac{\left(1 - {\cos}^{2} \left(\theta\right)\right) \left(1 + {\cos}^{2} \left(\theta\right)\right)}{\cos} ^ 2 \left(\theta\right) - \frac{1}{\cos \left(\theta\right) \cdot \sin \left(\theta\right)}$
using that

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

we get

$\frac{1 - {\cos}^{4} \left(\theta\right)}{\cos} ^ 2 \left(\theta\right) - \frac{1}{\cos \left(\theta\right) \sin \left(\theta\right)}$

where

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