If $A, B, C, D$ are the angles of a cyclic quadrilateral taken in order, then
$cos(180^o + A) + cos(180^o -B) + cos(180^o -C) -sin(90^o -D)=$
If $A, B, C, D$ are the angles of a cyclic quadrilateral taken in order, then
$cos(180^o + A) + cos(180^o -B) + cos(180^o -C) -sin(90^o -D)=$
- A
$0$
- B
$1$
- C
$-1$
- D
None of these
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- [KVPY 2017]
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