Differentiating $f(x)=\frac{3(1-\sin x)}{2\cos x}$ I'm having trouble to solve the following derivative. I tried to apply the quotient rule, but I cannot get the same result as my textbook.
The function:
$$f(x)=\frac{3(1-\sin x)}{2\cos x}$$
The result I supposed to get:
$$f'(x)=\frac{3}{2}\sec x(\tan x-\sec x)$$
I have tried this by applying the quotient rule:
$$f(x) =\frac{3(1-\sin x)}{2\cos x}=\frac{3-3\sin x}{2\cos x}$$
\begin{align*}
f'(x) & =\frac{-3\cos x \cdot 2\cos x - (3- \sin x) \cdot -2\sin x}{2\cos^2x}\\
      & =\frac{-6\cos^2x-(-6\sin x+6\sin^2x)}{2\cos^2x}\\
      & =\frac{-6\cos^2x+6\sin x+6\sin^2x}{2\cos^2x}
\end{align*}
Is someone able to show me step by step how to get the above result?
Thank you in advance for all of your help!

$$f(x)=\frac{3(1-\sin x)}{2\cos x} =\frac {3}{2} (\frac {1-\sin x}{\cos x})$$
Apply the quotient rule $$ f'(x) =\frac {3}{2}( \frac {-\cos x (\cos x)+\sin x (1-\sin x)}{\cos ^2 x})=$$
$$\frac {3}{2}( \frac {\sin x -1}{\cos ^2 x})=$$
$$\frac {3}{2}( \frac {1}{\cos x} \frac {\sin x -1}{\cos x})=$$
$$\frac {3}{2} \sec x(\tan x - sec x)$$
