Question

evaluate cos(40-theta) + sin(50-theta) + cos^2 40 + cos^2 50/sin^2 40 + sin^2 50

Answer 1

$$\begin{gathered} \cos \left(40°-\mathrm{\theta }\right)-\sin \left(50°+\mathrm{\theta }\right)+\frac{{\cos }^{2}40°+{\cos }^{2}50°}{{\sin }^{2}40°+{\sin }^{2}50°} \\ =\cos \left[90°-\left(50°+\mathrm{\theta }\right)\right]-\sin \left(50°+\mathrm{\theta }\right)+\frac{{\cos }^2\left(90°-50°\right)+{\cos }^{2}50°}{{\sin }^2\left(90°-50°\right)+{\sin }^{2}50°} \\ =\sin \left(50°+\mathrm{\theta }\right)-\sin \left(50°+\mathrm{\theta }\right)+\frac{{\sin }^{2}50°+{\cos }^{2}50°}{{\cos }^{2}50°+{\sin }^{2}50°} \\ =0+\frac{1}{1}\left[\mathrm{as},\cos \left(90°-\mathrm{\theta }\right)=\sin \mathrm{\theta },\sin \left(90°-\mathrm{\theta }\right)=\cos \mathrm{\theta },{\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=1\right] \\ =1 \end{gathered}$$