If ABCD is a cyclic quad. and A, B, C, D are interior angles then prove that
SinA/2 +Sin B/2 =CosC/2+CosD/2
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heya,
in cyclic quad <A+<C=180°------1)
And similarly ,<B+<D=180°
now,from 1)
<A=180°-<C
sinA/2=sin(90°-c/2)
sinA/2=cosc/2-------3)
similarly ,
sinB/2=cosD/2.-------4)
adding 1)and 2) we get
sinA/2+sinB/2=cosC/2+cosD/2
hope it help you.
@rajukumar☺
in cyclic quad <A+<C=180°------1)
And similarly ,<B+<D=180°
now,from 1)
<A=180°-<C
sinA/2=sin(90°-c/2)
sinA/2=cosc/2-------3)
similarly ,
sinB/2=cosD/2.-------4)
adding 1)and 2) we get
sinA/2+sinB/2=cosC/2+cosD/2
hope it help you.
@rajukumar☺
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