If A + B + C = pi and cos A = cos B cos C,the | vedclass

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

Question

(B) Given $A + B + C = \pi$,so $A = \pi - (B + C)$.Substituting this into the given equation $\cos A = \cos B \cos C$:$\cos(\pi - (B + C)) = \cos B \c

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given $A + B + C = \pi$,so $A = \pi - (B + C)$.Substituting this into the given equation $\cos A = \cos B \cos C$:$\cos(\pi - (B + C)) = \cos B \cos C$Using the identity $\cos(\pi - 	heta) = -\cos 	heta$:$-\cos(B + C) = \cos B \cos C$Expanding $\cos(B + C)$:$-(\cos B \cos C - \sin B \sin C) = \cos B \cos C$$-\cos B \cos C + \sin B \sin C = \cos B \cos C$$\sin B \sin C = 2 \cos B \cos C$Dividing both sides by $\cos B \cos C$ (assuming $\cos B, \cos C 
eq 0$):$\frac{\sin B \sin C}{\cos B \cos C} = 2$$	an B 	an C = 2$.

If $A + B + C = \pi$ and $\cos A = \cos B \cos C$,then $\tan B \tan C$ is equal to

Similar Questions

The value of $\frac{(\tan 69^{\circ} + \tan 66^{\circ})}{(1 - \tan 69^{\circ} \tan 66^{\circ})}$ is

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

The value of $\frac{\tan 40^{\circ}+\tan 20^{\circ}}{1-\cot 70^{\circ} \cot 50^{\circ}}$ is equal to

Evaluate: $\tan A + \cot (180^\circ + A) + \cot (90^\circ + A) + \cot (360^\circ - A)$

The maximum value of $\cos^2 \left( \frac{\pi}{3} - x \right) - \cos^2 \left( \frac{\pi}{3} + x \right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module