A) 1
B) - 1
C) 0
D) \[\frac{1}{2}\]
Correct Answer: C
Solution :
Since A, B, C, D are the angles of a quadrilateral, \[{{A}^{o}}+{{B}^{o}}+{{C}^{o}}+{{D}^{o}}=2\pi \] \[\therefore \] \[\sin (A+B)+\sin (C+D)\] \[=2\sin \,\frac{A+B+C+D}{2}\,\cos \frac{A+B-C-D}{2}\] \[=2\,\sin \pi \cos \frac{A+B-C-D}{2}\] \[=2\times 0\times \cos \frac{A+B-C-D}{2}=0\]You need to login to perform this action.
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