Solution

Once again, I will utilize the method of all Sine/Cosine to simplify: \[ \solve{ \dfrac{\sec x - \cos x}{\tan x}&=&\dfrac{\frac{1}{\cos x}- \cos x}{\frac{\sin x}{\cos x}}\\ &=&\left(\dfrac{1}{\cos x}-\dfrac{\cos^2x}{\cos x}\right)\times \dfrac{\cos x}{\sin x}\\ &=&\dfrac{1-\cos^2x}{\cos x}\times \dfrac{\cos x}{\sin x}\\ &=&\dfrac{\sin^2x}{\sin x}\\ &=&\sin x } \] The main big step, \(1-\cos^2x=\sin^2x\), is a direct result of the Pythagorean Identity: \(\cos^2 x +\sin^2x = 1\). The rest is all a result of canceling common terms and dividing by a fraction.