# What is the slope of the polar curve f(theta) = sectheta - csctheta  at theta = (3pi)/8?

Aug 3, 2017

$m = 6.757$

#### Explanation:

The slope of the tangent line of a function at a point is equal to the derivative of the function at that point.

With that being said, let's take the derivative

$\frac{\mathrm{df}}{d \theta} \left[f \left(\theta\right) = \sec \theta - \csc \theta\right]$

The derivative of $\sec \theta$ is $\tan \theta \sec \theta$:

$f ' \left(\theta\right) = \tan \theta \sec \theta - \frac{d}{d \theta} \left[\csc \theta\right]$

The derivative of $\csc \theta$ is $- \cot \theta \csc \theta$:

ul(f'(theta) = tanthetasectheta + cotthetacsctheta

Or, in terms of $\sin$ and $\cos$:

ul(f'(theta) = (sintheta)/(cos^2theta) + (costheta)/(sin^2theta)

Now, to find the slope at the point $\frac{3 \pi}{8}$, we plug it in for $\theta$:

m = (sin((3pi)/8))/(cos^2((3pi)/8)) + (cos((3pi)/8))/(sin^2((3pi)/8)) = color(blue)(2sqrt(10+sqrt2)

= color(blue)(ulbar(|stackrel(" ")(" "6.757" ")|)