Question

If ABCD is a cyclic quadrilateral, then:
\[\cos \left( {{180}^{\circ }}+A \right)+\cos \left( {{180}^{\circ }}-B \right)+\cos \left( {{180}^{\circ }}-C \right)-\sin \left( {{90}^{\circ }}-D \right)=\]
A. -1
B. 0
C. 1
D. 2

Answer 1

Hint: For the above question, we will have to know about a cyclic quadrilateral. As the name of a cyclic quadrilateral suggests, it’s a quadrilateral which is inscribed inside a circle. All of its vertices lie on that single circle. We will use the property of a cyclic quadrilateral that the sum of each pair of opposite angles is 180 degrees \[\left( {{180}^{\circ }} \right)\].

Complete step-by-step answer:
We have been given a cyclic quadrilateral ABCD and asked to find the value of the expression as follows:
\[\cos \left( {{180}^{\circ }}+A \right)+\cos \left( {{180}^{\circ }}-B \right)+\cos \left( {{180}^{\circ }}-C \right)-\sin \left( {{90}^{\circ }}-D \right)\]

Now as we know that the sum of each pair of opposite angles in a cyclic quadrilateral is equal to \[{{180}^{\circ }}\].
\[\begin{align}
  & \Rightarrow A+C={{180}^{\circ }} \\
 & \Rightarrow A={{180}^{\circ }}-C....(1) \\
\end{align}\]
Also,
\[\begin{align}
  & \Rightarrow B+D={{180}^{\circ }} \\
 & \Rightarrow D={{180}^{\circ }}-B....(2) \\
\end{align}\]
Now,
\[\cos \left( {{180}^{\circ }}+A \right)+\cos \left( {{180}^{\circ }}-B \right)+\cos \left( {{180}^{\circ }}-C \right)-\sin \left( {{90}^{\circ }}-D \right)\]
Since we know that \[\cos \left( {{180}^{\circ }}\pm \theta \right)=-\cos \theta \] and \[\sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta \],
\[\Rightarrow -\cos A-\cos B+\cos \left( {{180}^{\circ }}-C \right)-\cos D\]
On substituting the value of \[\left( {{180}^{\circ }}-C \right)=A\] and \[\left( {{180}^{\circ }}-B \right)=D\] in the above expression, we get as follows:
Since \[\cos \left( {{180}^{\circ }}-\theta \right)=-\cos \theta \],
\[\begin{align}
  & \Rightarrow -\cos A-\cos B+\cos A-\cos \left( {{180}^{\circ }}-B \right) \\
 & \Rightarrow -\cos A-\cos B+\cos A-\left( -\cos B \right) \\
\end{align}\]
On rearranging the terms, we get as follows:
\[\begin{align}
  & \Rightarrow -\cos A+\cos A-\cos B+\cos B \\
 & \Rightarrow 0 \\
\end{align}\]
Hence the value of the given expression is 0.
Therefore, the correct answer of the question is option B.

Note: Be careful while using the trigonometric identity like \[\cos \left( {{180}^{\circ }}-\theta \right)=-\cos \theta \] and also take care of the sign while calculating. In the given question, we assumed that B is less than equal to 90 degrees as the formula \[\sin \left( {{90}^{\circ }}-\theta \right)=\sin \theta \] is valid for \[\theta \le {{90}^{\circ }}\]. Also, remember the fact that the circle on which the cyclic quadrilateral lies is called the circumcircle or circumscribed circle.