Gujarati
Hindi
Trigonometrical Equations
normal

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

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