# What is the slope of the polar curve f(theta) = sectheta - csctheta  at theta = (3pi)/4?

May 3, 2018

Slope of the polar curve at $\theta = \frac{3 \pi}{4}$ is $0$

#### Explanation:

f(theta)=sec theta - csc theta ; theta = (3pi)/4

$f ' \left(\theta\right) = \sec \theta \cdot \tan \theta - \left(- \csc \theta \cdot \cot \theta\right)$ or

$f ' \left(\theta\right) = \sec \theta \cdot \tan \theta + \csc \theta \cdot \cot \theta$

$\cos \left(\frac{3 \pi}{4}\right) = - \frac{1}{\sqrt{2}} \therefore \sec \left(\frac{3 \pi}{4}\right) = - \sqrt{2}$

$\sin \left(\frac{3 \pi}{4}\right) = \frac{1}{\sqrt{2}} \therefore \csc \left(\frac{3 \pi}{4}\right) = \sqrt{2}$

$\tan \left(\frac{3 \pi}{4}\right) = - 1 \therefore \cot \left(\frac{3 \pi}{4}\right) = - 1$

$\therefore f ' \left(\theta\right) = \left(- \sqrt{2} \cdot - 1\right) + \left(\sqrt{2} \cdot - 1\right)$

$\therefore f ' \left(\theta\right) = \sqrt{2} - \sqrt{2} = 0$

Slope of the polar curve at $\theta = \frac{3 \pi}{4}$ is $0$ [Ans]