Question

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

• A. $0$
• B. $1$
• C. $2(\cos B–\cos D)$
• D. $2(\cos A–\cos C)$

Answer 1

The correct option is A

$0$

Explanation for the correct option:

Find the value of $\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{A}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{B}\mathbf{+}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{C}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{D}\mathbf{:}$

Given that $ABCD$ is a cyclic quadrilateral.

We know that the sum of opposite angles of a cyclic quadrilateral is equal to ${180}^{\circ }$.

$\begin{array}{rl}A+C& =180°\end{array}$

$\begin{array}{rl}B+D& =180°\end{array}$

$\begin{array}{rl}\cos A–\cos B+\cos C–\cos D& =\cos \left(180°–C\right)–\cos \left(180°–D\right)+\cos C–\cos D\\ & =–\cos C–(-\cos D)+\cos C–\cos D\\ & =0\end{array}$$\left[\mathbf{\because }\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{(}\mathbf{180}\mathbf{-}\mathbf{\theta }\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{\theta }\right]$

Hence option $\left(A\right)$ is the correct option.