Question

The value of $\cos ({40}^o+\theta )-\sin ({50}^o-\theta )+\frac{{\cos }^2{40}^o+{\cos }^2{50}^o}{{\sin }^2{40}^o+{\sin }^2{50}^o}$ is :

• A. $1$
• B. $2$
• C. $0$
• D. $3$

Answer 1

Answer: (A)

$1$

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

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

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

$=0+1=1$