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

If $\tan \theta + \cot \theta = 2$,then the value of $\tan^{100} \theta + \cot^{100} \theta$ is

If $180^{\circ} < \theta < 270^{\circ},$ then the value of $\sqrt{4 \sin^{4} \theta + \sin^{2} 2\theta} + 4 \cos^{2} \left(\frac{\pi}{4} - \frac{\theta}{2}\right)$ is

$\frac{\sin 70^\circ + \cos 40^\circ}{\cos 70^\circ + \sin 40^\circ} = $

$\frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}$ is equal to

If $k = \sin \frac{\pi}{18} \cdot \sin \frac{5\pi}{18} \cdot \sin \frac{7\pi}{18}$,then the numerical value of $k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module