Question

If $A={45}^{\circ },$ then the value of ${\cos }^{2}B+{\sin }^2(A+B)+2\sin A\sin ({180}^{\circ }+B)\cos ({360}^{\circ }+A+B)$ is

• A. $0$
• B. $\frac{1}{2}$
• C. $\frac{3}{2}$
• D. $2$

Answer 1

The correct option is C $\frac{3}{2}$

${\cos }^{2}B+{\sin }^2(A+B)+2\sin A\sin ({180}^{\circ }+B)\cos ({360}^{\circ }+A+B)$
We know that
$2\sin A\sin ({180}^{\circ }+B)\cos ({360}^{\circ }+A+B)=-2\sin A\sin B\cos (A+B)=-\left[\cos \right(A-B)-\cos (A+B\left)\right]\cos (A+B)=-\left[\cos \right(A+B\left)\cos \right(A-B)-{\cos }^2(A+B\left)\right]=-{\cos }^{2}A+{\sin }^{2}B+{\cos }^2(A+B)$

Now,
${\cos }^{2}B+{\sin }^2(A+B)+2\sin A\sin ({180}^{\circ }+B)\cos ({360}^{\circ }+A+B)={\cos }^{2}B+{\sin }^2(A+B)-{\cos }^{2}A+{\sin }^{2}B+{\cos }^2(A+B)=2-{\cos }^{2}A=2-\frac{1}{2}=\frac{3}{2}$