sir help

Asked by vt2008 | 19th Sep, 2010, 05:02: PM

Expert Answer:

we know that sum of the opp angles of a cyclic quadrilateral are supplementary,
so
A+C=180 degrees
B+D=180 degrees
cos(180-A)=cos C
cos (180+B)=-cos B...as cos(180+x)=-cos x
cos(180+C)=-cosC
sin (90 +D)=cos D=cos (180-B)=-cos B...using the formula sin(90 +x)=cos x
So we get,
=0
LHS=cos C+(-cos B)+(-cos C)+(-cos B)
=0
=RHS.Hence proved

Answered by  | 19th Sep, 2010, 09:45: PM

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