# What is the slope of the tangent line of secx/cscy= C , where C is an arbitrary constant, at (pi/3,pi/3)?

Nov 11, 2016

I would rewrite and find $C$ first.

#### Explanation:

$\sec \frac{x}{\csc} y = \sin \frac{y}{\cos} x = C$

Given that the point $\left(\frac{\pi}{3} , \frac{\pi}{3}\right)$ lies on the graph, we find $C = \sqrt{3}$

Therefore,

$\sin y = \sqrt{3} \cos x$.

Differentiating gets us

$\cos y \frac{\mathrm{dy}}{\mathrm{dx}} = - \sqrt{3} \sin x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \sqrt{3} \sin \frac{x}{\cos} y$.

At $\left(\frac{\pi}{3} , \frac{\pi}{3}\right)$ then, we have

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \sqrt{3} \sin \frac{\frac{\pi}{3}}{\cos} \left(\frac{\pi}{3}\right) = - 3$