Trigonometry – II — Maths STD 11 Science — Question
Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II2 Marks
MCQ
If ABCD is a cyclic quadrilateral, then $\cos A +\cos B$ is equal to
A
$0$
B
$\cos C+\cos D$
✓
$-(\cos C+\cos D)$
D
$\cos C-\cos D$
✓
Answer
Correct option: C.
$-(\cos C+\cos D)$
(C) Since $A+C=180^{\circ}$ and $B+D=180^{\circ}$ $\therefore \cos A+\cos B=\cos \left(180^{\circ}-C\right)+\cos \left(180^{\circ}-D\right)$ $=-(\cos C +\cos D )$ $\ldots\left[\because \cos \left(180^{\circ}-\theta\right)=-\cos \theta\right]$
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