Question

$\sin {43}^{\circ }\cos {47}^{\circ }+\cos {43}^{\circ }\sin {47}^{\circ }=$

• A. $0$
• B. $1$
• C. $\sin 4^{\circ }$
• D. $\cos 4^{\circ }$

Answer 1

The correct option is B $1$

Taking

$\sin \mathring{43}\cos \mathring{47}+\cos \mathring{43}\sin \mathring{47}$

Writing sin $47$ and cos$47$ in form of angle $43$ so we get

$=\sin \mathring{43}\cos \left(\mathring{90-}\mathring{43}\right)+\cos \mathring{43}\sin \left(\mathring{90-}\mathring{43}\right)$

We know that sin (90-θ)= cos θ and cos (90-θ)=sin θ

$=\sin \mathring{43}\sin \mathring{43}+\cos \mathring{43}\cos \mathring{43}$

$={\sin }^2\mathring{43}+{\cos }^2\mathring{43}$

$=1$