Question

The value of sin$({60}^{\circ }+\theta )-$ cos$({30}^{\circ }-\theta )$ is equal to

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

Answer 1

The correct option is C $0$

$\sin (60°+\theta )-\cos (30°-\theta )$

$=\cos [90°-(60°+\theta )]-\cos (30°-\theta )$

$=\cos (30°-\theta )-\cos (30°-\theta )$

$=0$

Hence, the answer is $0.$