What is value of \(\frac{d}{dx}(\frac{cos⁡x}{sec⁡x \,tan⁡x})\)?

\(\frac{d}{dx}(\frac{cos?x}{sec?x \,tan?x}) = \frac{d}{dx}(\frac{cos?x}{\frac{1}{cos?x}\frac{sin?x}{cos?x}})\)
=\(\frac{d}{dx}(\frac{cos^3?x}{sin?x})\)
Using quotient rule, we know that, \(\frac{d}{dx} (\frac{f}{g}) = \frac{g.\frac{d}{dx} (f) – f.\frac{d}{dx}(g)}{g^2}\)
Here, f = cos³? x and g = sin ?x
\(\frac{d}{dx}(\frac{cos^3?x}{sin?x})\) = \(\frac{sin?x.\frac{d}{dx} (cos^3?x)-cos^3?x.\frac{d}{dx}(sinx)}{sin^2?x}\)
\(\frac{d}{dx}(\frac{cos^3?x}{sin?x})\) = \(\frac{sin?x (-3 cos^2? x \,sin? x) – cos^3?x(cos?x)}{sin^2?x}\)
\(\frac{d}{dx}(\frac{cos^3?x}{sin?x})\) = \(\frac{-3 sin^2?x \,cos^2?x – cos^4?x}{sin^2?x}\)
\(\frac{d}{dx}(\frac{cos^3?x}{sin?x})\) = -cot²? x(3 sin²? x + cos² x)