Докажите тождество ctg(70)-sqrt(3)=-4cos(70)

$ctg 70^\circ - \sqrt{3} =-4cos70^\circ$

$ctg70^\circ -ctg30^\circ =-4cos70^\circ$

$\frac{sin(30^\circ -70^\circ )}{sin70^\circ sin30^\circ } =-4cos70^\circ$

$\frac{-sin40^\circ }{sin70* \frac{1}{2} }=-4 cos70^\circ$

$\frac{-2sin40^\circ }{sin70^\circ } =-4cos70^\circ$

$\frac{-2*2sin20^\circ cos20^\circ }{sin(90^\circ -20^\circ )} =-4cos70^\circ$

$\frac{-4sin20^\circ cos20^\circ }{cos20^\circ } =-4cos70^\circ$

$-4sin20^\circ =-4cos70^\circ$

$-4sin(90^\circ -70^\circ )=-4cos70^\circ$

$-4cos70^\circ =-4cos70^\circ$

что и требовалось доказать