Вычислить производную f(x) = tgx-4ctgxp.s. без sec cosec

Решите задачу:

$f(x)=tg(x)-4ctg(x);\\ tg(x)= \frac{\sin(x)}{\cos(x)};\ \ ctg(x)=(tg(x))^{-1}= \frac{\cos(x)}{\sin(x)};\\ \ \ (f\pm g)'=f'\pm g'; ( \frac{f}{g} )'= \frac{f'\cdot g-f\cdot g'}{g^2}\\ \sin'(x)=\cos(x);\ \ \cos'(x)=-\sin(x)\\ f'(x)=(tg(x)-4ctg(x))'=tg'(x)-4ctg'(x)= (\frac{\sin(x)}{\cos(x)})'-4 (\frac{\cos(x)}{\sin(x)})'=\\ = \frac{\sin'(x)\cos(x)-\sin(x)\cos'(x)}{\cos^2(x)} -4 \frac{\cos'(x)sin(x)-\cos(x)\sin'(x)}{\sin^2(x)}=\\$
$= \frac{\cos^2(x)+\sin^2(x)}{cos^2(x)}-4 \frac{-sin^2(x)-\cos^2(x)}{sin^2(x)}=\\ = \frac{1}{\cos^2(x)}+ \frac{4}{\sin^2(x)}= \frac{\sin^2(x)+4\cos^2(x)}{\sin^2(x)\cos^2(x)} \\ =\frac{1+3\cos^2(x)}{ \frac{1}{4} \sin^2(2x)}= \frac{4(1+3\cos^2(x))}{\sin^2(2x)}$