If $$\cos (A + B) = 0$$ and $$\sin (A - B) = \frac{1}{2}$$ , then the value of B is:

(Given $$0^\circ < A, B < 90^\circ$$)

Given,  $$0^\circ < A, B < 90^\circ$$

$$\cos (A + B) = 0$$

$$=$$>  $$\cos(A+B)=\cos90^{\circ\ }$$

$$=$$>  $$A+B=90^{\circ\ }$$ .....................(1)

$$\sin (A - B) = \frac{1}{2}$$

$$=$$>  $$\sin(A-B)=\sin30^{\circ\ }$$

$$=$$>  $$A-B=30^{\circ\ }$$ .....................(2)

Subtracting (2) from (1), we get

$$A+B-A+B=90^{\circ\ }-30^{\circ\ }$$

$$=$$>  $$2B=60^{\circ\ }$$

$$=$$>  $$B=30^{\circ\ }$$

Hence, the correct answer is Option D