Question

Assertion :If $A,B,C,D$ are angles of a cyclic quadrilateral then $\sum \sin A=0$ Reason: If $A,B,C,D$ are angles of a cyclic quadrilateral then $\sum \cos A=0$

• A. Both Assertion and Reason are individually true and Reason is the correct explanation of Assertion.
• B. Both Assertion and Reason are individually correct but Reason is not the correct explanation of Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is D Assertion is incorrect but Reason is correct

Since $ABCD$ is a cyclic quadrilateral
$\therefore A+C={180}^0,B+D={180}^0$
$\Rightarrow A={180}^0-C,B={180}^0-D$
$\Rightarrow \cos A=\cos ({180}^0-C),\cos B=\cos ({180}^0-D)$
$\Rightarrow \cos A=-\cos C,\cos B=-\cos D$
$\Rightarrow \cos A+\cos C=0,\cos B+\cos D=0$
Now, $\sum \cos A=\cos A+\cos B+\cos C+\cos D$
$=(\cos A+\cos C)+(\cos B+\cos D)$
$=0+0=0$
$\therefore \sum \cos A=0$
and again $\sin A=\sin ({180}^0-C),\sin B=\sin ({180}^0-D)$
$\Rightarrow \sin A=\sin C,\sin B=\sin D$
Now, $\sum \sin A=\sin A+\sin B+\sin C+\sin D$
$=2(\sin A+\sin B)\ne 0$
$=2(\sin C+\sin D)\ne 0$