# Question 297ca

Jul 10, 2017

$\cot \theta$

#### Explanation:

$\text{using the "color(blue)"trigonometric identities}$

•color(white)(x)cottheta=costheta/sintheta

•color(white)(x)sectheta=1/costheta#

$\Rightarrow {\cot}^{2} \theta . \sin \theta . \sec \theta$

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

$= \frac{\cancel{\cos \theta} \cos \theta}{\cancel{\sin \theta} \sin \theta} \times \cancel{\sin \theta} \times \frac{1}{\cancel{\cos \theta}}$

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

$= \cot \theta$

$\textcolor{red}{\text{OR}}$

$\cos \frac{\theta}{\sin} \theta = \cos \theta \times \frac{1}{\sin} \theta = \cos \theta \csc \theta$

Jul 10, 2017

${\cot}^{2} \theta \cdot \sin \theta \cdot \sec \theta = \cot \theta$

#### Explanation:

${\cot}^{2} \theta \cdot \sin \theta \cdot \sec \theta$

Rewrite in terms of $\sin$ and $\cos$

$= \left({\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta\right) \cdot \sin \theta \cdot \left(\frac{1}{\cos} \theta\right)$

$= \frac{\cos \theta \cdot \textcolor{red}{\cos \theta} \cdot \textcolor{red}{\sin \theta}}{\sin \theta \cdot \textcolor{red}{\sin \theta} \cdot \textcolor{red}{\cos \theta}}$

$= \cot \theta$