# How do you express 4 cos theta - sec theta + 2 cot theta  in terms of sin theta ?

Mar 18, 2016

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

#### Explanation:

We shall use the following identities:

$\sec \theta = \frac{1}{\cos} \theta$

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

Also, ${\cos}^{2} \theta + {\sin}^{2} \theta = 1$

$\implies {\cos}^{2} \theta = 1 - {\sin}^{2} \theta \textcolor{red}{\rightarrow} \cos \theta = \pm \sqrt{1 - {\sin}^{2} \theta}$

The $+$ or $-$ depends on the quadrant in which the angle $\theta$
is found.

Assuming $\theta$ acute we can simply ignore this.

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

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

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

$\implies 4 \cos \theta - \sec \theta + 2 \cot \theta \textcolor{red}{\rightarrow} 4 \cdot \left(\sqrt{1 - {\sin}^{2} \theta}\right) - \left(\frac{1}{\sqrt{1 - {\sin}^{2} \theta}}\right) + 2 \cdot \left(\frac{\sqrt{1 - {\sin}^{2} \theta}}{\sin} \theta\right)$

$\textcolor{red}{\rightarrow} \frac{4 \left(1 - {\sin}^{2} \theta\right) \sin \theta - \sin \theta + 2 \left(1 - {\sin}^{2} \theta\right)}{\sin \theta \sqrt{1 - {\sin}^{2} \theta}}$