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Question: If ABCD is a cyclic quadrilateral, then the value of \(\cos A - \cos B + \cos C - \cos D =\)...

If ABCD is a cyclic quadrilateral, then the value of cos⁡A−cos⁡B+cos⁡C−cos⁡D=\cos A - \cos B + \cos C - \cos D =

A

0

B

1

C

2(cos⁡B−cos⁡D)2(\cos B - \cos D)

D

2(cos⁡A−cos⁡C)2(\cos A - \cos C)

Answer

0

Explanation

Solution

We know that A+C=180∘,A + C = 180{^\circ},since ABCDABCD is a cyclic

quadrilateral.⇒A=180∘−C\Rightarrow A = 180{^\circ} - C

⇒cos⁡A=cos⁡(180∘−C)=−cos⁡C\Rightarrow \cos A = \cos(180{^\circ} - C) = - \cos C

⇒cos⁡A+cos⁡C=0\Rightarrow \cos A + \cos C = 0 .....(i)

Now B+D=180∘,B + D = 180{^\circ},then cos⁡B+cos⁡D=0\cos B + \cos D = 0 .....(ii)

Subtracting (ii) from (i), we get

cos⁡A−cos⁡B+cos⁡C−cos⁡D=0\cos A - \cos B + \cos C - \cos D = 0.