Question

Evaluate :

$\cos ({40}^o+\theta )-\sin ({50}^o-\theta )+\cos {240}^o+\cos {250}^o\sin {240}^o+\sin {250}^o=?$

Answer 1

Use concept $\cos ({90}^o-\theta )=\sin \theta$

and $\sin ({90}^o-\theta )=\cos \theta$

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

$=\cos ({40}^o+\theta )-\sin ({50}^o-\theta )+\frac{{\cos }^2({90}^o-{50}^o)+{\cos }^2({90}^o-{40}^o)}{{\sin }^2{40}^o+{\sin }^2{50}^o}$

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

$=\cos ({40}^o+\theta )-\sin [{90}^o-({40}^o+\theta \left)\right]+1$

$=\cos ({40}^o+\theta )-\cos ({40}^o+\theta )+1=0+1=1$.