Question

Solve:
$\cos (40-\theta )-\sin (50+\theta )+\frac{{\cos }^{2}40+{\cos }^{2}50}{{\sin }^{2}40+{\sin }^{2}50}$

Answer 1

$\cos (40-\theta )-\sin (50+\theta )+\frac{{\cos }^{2}40+{\cos }^{2}50}{{\sin }^{2}40+{\sin }^{2}50}$

$=\cos (40-\theta )-\sin (90-(40-\theta ))+\frac{{\cos }^{2}40+{\cos }^2(90-40)}{{\sin }^{2}40+{\sin }^2(90-40)}$

$=\cos (40-\theta )-\cos (40-\theta )+\frac{{\cos }^{2}40+{\sin }^{2}40}{{\sin }^{2}40+{\cos }^{2}40}$

$=0+\frac{1}{1}$

$=1$