If A + B + C = 180^circ, then frac{sin 2A + s | vedclass

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

Question

(B) Given $A + B + C = 180^\circ.$Numerator $(N^r)$ = $\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C = 4(2\sin \frac{A}{2}\cos \frac{A}{2})(2\si

Vedclass · Accepted Answer

$8\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}$

Vedclass · Answer

(B) Given $A + B + C = 180^\circ.$Numerator $(N^r)$ = $\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C = 4(2\sin \frac{A}{2}\cos \frac{A}{2})(2\sin \frac{B}{2}\cos \frac{B}{2})(2\sin \frac{C}{2}\cos \frac{C}{2}) = 32\sin \frac{A}{2}\cos \frac{A}{2}\sin \frac{B}{2}\cos \frac{B}{2}\sin \frac{C}{2}\cos \frac{C}{2}.$Denominator $(D^r)$ = $\cos A + \cos B + \cos C - 1 = -4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}.$Wait,using the identity $\cos A + \cos B + \cos C = 1 + 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2},$ we have $D^r = 4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}.$Now,$N^r = 4\sin A \sin B \sin C = 4(2\sin \frac{A}{2}\cos \frac{A}{2})(2\sin \frac{B}{2}\cos \frac{B}{2})(2\sin \frac{C}{2}\cos \frac{C}{2}) = 32\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}.$Therefore,$\frac{N^r}{D^r} = \frac{32\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}}{4\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}} = 8\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2}.$

If $A + B + C = 180^\circ,$ then $\frac{\sin 2A + \sin 2B + \sin 2C}{\cos A + \cos B + \cos C - 1} = $

Similar Questions

If $\cos(\alpha + \beta) = \frac{3}{5}$,$\sin(\alpha - \beta) = \frac{5}{13}$ and $0 < \alpha, \beta < \frac{\pi}{4}$,then $\tan(2\alpha)$ is equal to

$\left( \frac{\sin 2A}{1 + \cos 2A} \right) \left( \frac{\cos A}{1 + \cos A} \right) = $

If $\sin (\theta + \alpha ) = a$ and $\sin (\theta + \beta ) = b,$ then $\cos 2(\alpha - \beta ) - 4ab\cos (\alpha - \beta )$ is equal to

$\frac{{\cos 12^\circ - \sin 12^\circ }}{{\cos 12^\circ + \sin 12^\circ }} + \frac{{\sin 147^\circ }}{{\cos 147^\circ }} = $

If $x \sin^{2} 60^{\circ} - \frac{3}{2} \sec 60^{\circ} \tan^{2} 30^{\circ} + \frac{4}{5} \sin^{2} 45^{\circ} \tan^{2} 60^{\circ} = 0$,then $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module