Question

$\cos ({40}^{\circ }+\theta )-\sin ({50}^{\circ }-\theta )=$

• A. $1$
• B. $0$
• C. $\sin {20}^{\circ }$
• D. None of these

Answer 1

Answer: (B)

$0$

Given

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

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

$=\cos \left({40}^o+\theta \right)-\cos \left({40}^o+\theta \right)$ as we know $\sin ({90}^o-\theta )=\cos \theta$

$=0$