Question

$\sin ({45}^{\circ }+\theta )-\cos ({45}^{\circ }-\theta )$ is equal to

• A. $2\cos \theta$
• B. $0$
• C. $2\sin \theta$
• D. $1$

Answer 1

The correct option is B $0$

$\sin ({45}^0+\theta )-\cos ({45}^0-\theta )$
$=\cos ({90}^0-({45}^0+\theta \left)\right)-\cos ({45}^0-\theta )$
$=\cos ({45}^0-\theta )-\cos ({45}^0-\theta )$
$=0$