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Trigonometrical Equations
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If $A, B, C, D$ are the angles of a cyclic quadrilateral taken in order, then
$cos(180^o + A) + cos(180^o -B) + cos(180^o -C) -sin(90^o -D)=$
A
$0$
B
$1$
C
$-1$
D
None of these
Solution
Given expression
$=-\cos A-\cos B-\cos C-\cos D$
$=-[(\cos A+\cos C)+(\cos B+\cos D)]$
$=-(0+0)=0$
$[\because \mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ are the angles of a cyclic quadrilateral $\left.\therefore A+C=180^{\circ} \Rightarrow B+D=180^{\circ}\right]$
Standard 11
Mathematics