Question

$\sin ({45}^0+\theta )-\cos ({45}^0-\theta )$ is equal to

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

Answer 1

Answer: (B)

$0$

$\sin ({45}^0+\theta )-\cos ({45}^0-\theta )$

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

$=sin({45}^0+\theta )-sin({45}^0+\theta )$

$=0$