If any quadrilateral ABCD, prove that `"sin"(A+B)+sin(C+D)=0` `"cos"(A+B)=cos(C+D)`

Explanation

In a quadrilateral `ABCD`,

`A+B+C+D = 2pi`

`=>A+B = 2pi-(C+D)`

`:. sin(A+B)+sin(C+D) = sin(2pi-(C+D))+sin(C+D)`

As, `sin(2pi-theta) = -sintheta`

`:. sin(2pi-(C+D))+sin(C+D) = -sin(C+D)+sin(C+D) = 0`

`:. sin(A+B)+sin(C+D) = 0`

Now,

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