Question

Without using trigonometric tables, evaluate that :
$cos({40}^{\circ }+\theta )-sin({50}^{\circ }-\theta )+\frac{co{s}^2{40}^{\circ }+co{s}^2{50}^{\circ }}{si{n}^2{40}^{\circ }+si{n}^2{50}^{\circ }}$

Answer 1

$\cos (40°+\theta )=\sin (90°-(40°+\theta ))=\sin (50°-\theta )$

$\cos (40°+\theta )-\sin (50°-\theta )=0$

$\cos 40°=\sin (90°-40°)=\sin 50°$

$\sin 40°=\cos (90°-40°)=\cos 50°$

$\frac{{\cos }^{2}40°+{\cos }^{2}50°}{{\sin }^{2}40°+{\sin }^{2}50°}=\frac{{\cos }^{2}40°+{\sin }^{2}40°}{{\sin }^{2}40°+{\cos }^{2}40°}=1$

$\therefore \cos (40°+\theta )-\sin (50°-\theta )+\frac{{\cos }^{2}40°+{\cos }^{2}50°}{{\sin }^{2}40°+{\sin }^{2}50°}=0+1=1$