# What is the equation of the tangent line of f(x)=cotx*secx at x=pi/4?

Dec 22, 2015

$y - \sqrt{2} = \sqrt{2} \left(x - \frac{\pi}{4}\right)$

#### Explanation:

First, simplify $f \left(x\right)$.

$f \left(x\right) = \cos \frac{x}{\sin} x \left(\frac{1}{\cos} x\right) = \frac{1}{\sin} x = \csc x$

Thus, $\frac{d}{\mathrm{dx}} \left[\csc x\right] = - \csc x \cot x$

This is identity you should know. It can be proven through the qoutient rule $\frac{d}{\mathrm{dx}} \left[\frac{1}{\sin} x\right]$ or the chain rule ${\left(\sin x\right)}^{-} 1$.

Since $f ' \left(x\right) = - \csc x \cot x$, we know that:

$f ' \left(\frac{\pi}{4}\right) = - \cot \left(\frac{\pi}{4}\right) \csc \left(\frac{\pi}{4}\right) = - 1 \left(\sqrt{2}\right) = - \sqrt{2}$

We also know that

$f \left(\frac{\pi}{4}\right) = \sqrt{2}$

Thus, the tangent line passes through the point $\left(\frac{\pi}{4} , \sqrt{2}\right)$ and has a slope of $- \sqrt{2}$.

Write this in point-slope form:

$y - \sqrt{2} = - \sqrt{2} \left(x - \frac{\pi}{4}\right)$