Question

lf $A,B,C,D$ are four angles then $\sin (A-B).\cos (A+B)+\sin (B-C)$. $\cos (B+C)+\sin (C-D).\cos (C+D)+\sin (D-A).\cos (D+A)=$

• A. $0$
• B. $1$
• C. $4\sin A.\sin B\sin C\sin D$
• D. $4\cos A\cos B\cos C\cos D$

Answer 1

The correct option is A $0$

Given $A,B,C,D$ are four angles then

$\sin (A-B).\cos (A+B)+\sin (B-C)\cos (B+C)+\sin (C-D).\cos (C+D)+\sin (D-A).\cos (D+A)$

multiply and divide by $2$

$\frac{1}{2}\left[2\sin \right(A-B\left)\cos \right(A+B)+2\sin (B-C\left)\cos \right(B+C)+2\sin (C-D\left)\cos \right(C+D)+2\sin (D-A\left)\cos \right(D+A\left)\right]$

$\frac{1}{2}(\sin A-\sin B+\sin B-\sin C+\sin C-\sin D+\sin D-\sin A)$

$=0$

Hence, option 'A' is correct.