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MHT CET · Maths · Trigonometric Ratios & Identities

If \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}\) are the angles of a cyclic quadrilateral taken in order, then \(\cos A+\cos B+\operatorname{Cos} C+\operatorname{Cos} D=\)

  1. A \(-1\)
  2. B 1
  3. C \(\frac{1}{2}\)
  4. D 0
Verified Solution

Answer & Solution

Correct Answer

(D) 0

Step-by-step Solution

Detailed explanation

Since the quadrilateral \(\mathrm{ABCD}\) is cyclic, we have
\(\mathrm{A}+\mathrm{C}=180^{\circ}\) and \(\mathrm{B}+\mathrm{D}=180^{\circ}\)
\(\therefore \cos \mathrm{A}=\cos \left(180^{\circ}-\mathrm{C}\right)=-\cos \mathrm{C}\)
\(\cos \mathrm{B}=\cos \left(180^{\circ}-\mathrm{D}\right) \quad=-\cos \mathrm{D}\)
\(\therefore \cos \mathrm{A}+\cos \mathrm{B}+\cos \mathrm{C}+\cos \mathrm{D}=0\)