Question

In a quadrilateral ABCD, prove that cos(A+B) = cos ( C + D) ​

Answer 1

Answer:

hi

Step-by-step explanation:

In a quadrilateral ABCD,

Solution :

A+B+C+D=2πA+B+C+D=2π

⇒A+B=2π−(C+D)⇒A+B=2π-(C+D)

∴sin(A+B)+sin(C+D)=sin(2π−(C+D))+sin(C+D)∴sin(A+B)+sin(C+D)=sin(2π-(C+D))+sin(C+D)

As, sin(2π−θ)=−sinθsin(2π-θ)=-sinθ

∴sin(2π−(C+D))+sin(C+D)=−sin(C+D)+sin(C+D)=0∴sin(2π-(C+D))+sin(C+D)=-sin(C+D)+sin(C+D)=0

∴sin(A+B)+sin(C+D)=0∴sin(A+B)+sin(C+D)=0

Now,

cos(A+B)=cos(2π−(C+D))cos(A+B)=cos(2π-(C+D))

As, cos(2π−θ)=cosθcos(2π-θ)=cosθ

∴cos(2π−(C+D))=cos(C+D)∴cos(2π-(C+D))=cos(C+D)

∴cos(A+B)=cos(C+D).

Hope it helps you

Answer 2

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