Question

if a,b,c,d are the angles of a quadrilateral, prove that cos1/2(a+b)+cos1/2(c+d)=0​

Answer 1

Step-by-step explanation:

We know that,

cos x+cos y,=2cos[(x+y)/2]* cos[(x-y)/2]

therefore,

=cos[A+B)/2]*cos[(C+D)/2]

=cos[(A+B+C+D)/4]* cos[(A+B-C-D)/4]

=2cos(360°/4)* cos[(A+B-C-D)/4]

(sum of the measures of all angles of a quadrilateral is 360 degree)

=2cos90°* cos[(A+B-C-D)/4]

=0* cos1/2(A+B)+cos1/2(C+D)

=0

LHS=0 AND RHS=0

HENCE,LHS=RHS PROVED.....

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