Question

Solve:
$\frac{\cos 35}{\sin 55}+\frac{\tan 27\tan 63}{\sin 30}-3(\tan 60{)}^2$

Answer 1

$\frac{\cos 35°}{\sin 35°}+\frac{tan27°tan63°}{\sin 30°}-3tan(60°{)}^2=\frac{\cos (90-55)°}{\sin 55°}+\frac{tan27°tan(90-27)°}{\frac{1}{2}}-3\times (\sqrt{3}{)}^2=\frac{\sin 55°}{\sin 55°}+tan27°cot27°\times 2-3\times 3=1+2-9=3-9=-6$