If ABCD is a cyclic quadrilateral,then the va | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) In a cyclic quadrilateral $ABCD$,the sum of opposite angles is $180^\circ$.Therefore,$A + C = 180^\circ$ and $B + D = 180^\circ$.From $A + C = 180

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) In a cyclic quadrilateral $ABCD$,the sum of opposite angles is $180^\circ$.Therefore,$A + C = 180^\circ$ and $B + D = 180^\circ$.From $A + C = 180^\circ$,we have $A = 180^\circ - C$.Taking cosine on both sides,$\cos A = \cos(180^\circ - C) = -\cos C$,which implies $\cos A + \cos C = 0$.Similarly,from $B + D = 180^\circ$,we have $B = 180^\circ - D$.Taking cosine on both sides,$\cos B = \cos(180^\circ - D) = -\cos D$,which implies $\cos B + \cos D = 0$.Now,consider the expression $\cos A - \cos B + \cos C - \cos D$.Rearranging the terms,we get $(\cos A + \cos C) - (\cos B + \cos D)$.Substituting the values derived above,we get $0 - 0 = 0$.

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

Similar Questions

Let $A_1 A_2 A_3 \ldots A_9$ be a nine-sided regular polygon with side length $2$ units. The difference between the lengths of the diagonals $A_1 A_5$ and $A_2 A_4$ equals

In a triangle $ABC$,if $A = \frac{\pi}{4}$ and $\tan B \tan C = K$,then $K$ must satisfy:

In $\triangle PQR$,if $\angle R = \frac{\pi}{4}$ and $\tan(\frac{P}{3})$,$\tan(\frac{Q}{3})$ are the roots of the equation $ax^2 + bx + c = 0$,then:

The common roots of the equations $2\sin^2 x + \sin^2 2x = 2$ and $\sin 2x + \cos 2x = \tan x$ are

If $\cos A + \cos B + 2\cos C = 2$,then the sides of the $\Delta ABC$ are in

Vedclass Test Series

Exam Paper Generator

Online Exam Module