Question

in any quadrilateral abcd prove that cos(a+b)=cos(c+d)​

Answer 1

Answer:

In a Quadrilateral ABCD,{∠A+∠B+∠C+∠D=360

∘

}

as ∠A+∠B=360

∘

−(∠C+∠D)

∴sin(A+B)=sin(360

∘

−(∠C+∠D))

=−sin(∠C+∠D) ___ (1)

∴sin(A+B)+sin(C+D)

⇒−sin(C+D)+sin(C+D)=0

(I) is true

(II) cos(A+B)=cos(C+D)=?

=cos(360

∘

−(A+B))

=cos(A+B)

∴cos(A+B)=cos(C+D)

Proved.

Answer 2

Answer:

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Asked on December 20, 2019 by

Barnita Harish

For any quadrilateral ABCD which of the following statement are true? sin(A+B)+sin(C+D)=0

cos(A+B)=cos(C+D)

Select the correct answer using the codes given below

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ANSWER

In a Quadrilateral ABCD,{∠A+∠B+∠C+∠D=360

∘

}

as ∠A+∠B=360

∘

−(∠C+∠D)

∴sin(A+B)=sin(360

∘

−(∠C+∠D))

=−sin(∠C+∠D) ___ (1)

∴sin(A+B)+sin(C+D)

⇒−sin(C+D)+sin(C+D)=0

(I) is true

(II) cos(A+B)=cos(C+D)=?

=cos(360

∘

−(A+B))

=cos(A+B)

∴cos(A+B)=cos(C+D)

Proved.

∴ (I) & II are true

(C) is correct.