Question

Question 14
$sin({45}^{\circ }+\theta )-cos({45}^{\circ }-\theta )$ is equal to

(A) $2cos\theta$
(B) 0​​​​​​​
(C) $2sin\theta$​​​​​​​
(D) 1​​​​​​​

Answer 1

The answer is (B).
$sin({45}^{\circ }+\theta )-cos({45}^{\circ }-\theta )=cos\lfloor {90}^{\circ }-({45}^{\circ }+\theta )\rfloor cos({45}^{\circ }-\theta )$

$\lfloor \because cos({90}^{\circ }-\theta )\rfloor =sin\theta$

$=cos({45}^{\circ }-\theta )-cos({45}^{\circ }-\theta )$

$=0$