Question

Solve:

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

Answer 1

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

$\left[\begin{array}{c}Weknowthat\\ \sin \left(90-\theta \right)=\cos \theta \\ \cos \left(90-\theta \right)=\sin \theta \end{array}\right]$

$\Rightarrow \sin \left[90-\left(40-\theta \right)\right]-\sin \left(50+\theta \right)+\frac{{\cos }^2\left(90-40\right)+{\cos }^{2}50}{{\sin }^2\left(90-40\right)+{\sin }^{2}50}$

$=\sin \left(50+\theta \right)-\sin \left(50+\theta \right)+\frac{{\sin }^{2}50+{\cos }^{2}50}{{\cos }^{2}50+{\sin }^{2}50}$

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