If cos 2B = frac{cos (A + C)}{cos (A - C)},th | vedclass

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

Question

(B) Given: $\cos 2B = \frac{\cos (A + C)}{\cos (A - C)}$Using the expansion formulas for $\cos(A+C)$ and $\cos(A-C)$:$\cos 2B = \frac{\cos A \cos C -

Vedclass · Accepted Answer

$G.P.$

Vedclass · Answer

(B) Given: $\cos 2B = \frac{\cos (A + C)}{\cos (A - C)}$Using the expansion formulas for $\cos(A+C)$ and $\cos(A-C)$:$\cos 2B = \frac{\cos A \cos C - \sin A \sin C}{\cos A \cos C + \sin A \sin C}$Divide the numerator and denominator by $\cos A \cos C$:$\cos 2B = \frac{1 - 	an A 	an C}{1 + 	an A 	an C}$We know that $\cos 2B = \frac{1 - 	an^2 B}{1 + 	an^2 B}$.Equating the two expressions:$\frac{1 - 	an^2 B}{1 + 	an^2 B} = \frac{1 - 	an A 	an C}{1 + 	an A 	an C}$By applying componendo and dividendo or cross-multiplying:$(1 - 	an^2 B)(1 + 	an A 	an C) = (1 + 	an^2 B)(1 - 	an A 	an C)$$1 + 	an A 	an C - 	an^2 B - 	an^2 B 	an A 	an C = 1 - 	an A 	an C + 	an^2 B - 	an^2 B 	an A 	an C$Simplifying the equation:$2 	an A 	an C = 2 	an^2 B$$	an^2 B = 	an A 	an C$Since the square of the middle term is equal to the product of the first and third terms,$	an A, 	an B, 	an C$ are in $G.P.$

If $\cos 2B = \frac{\cos (A + C)}{\cos (A - C)}$,then $\tan A, \tan B, \tan C$ are in

Similar Questions

If $\tan 20^{\circ} = k,$ then $\frac{\tan 250^{\circ} + \tan 340^{\circ}}{\tan 200^{\circ} - \tan 110^{\circ}} = $

If $A + B + C = \frac{3\pi}{2},$ then $\cos 2A + \cos 2B + \cos 2C = $

If $\frac{\cos 3A + \sin 3A}{\cos A - \sin A} = 1 - K \sin 2A,$ the value of $K$ is

$\cos \alpha \sin (\beta - \gamma ) + \cos \beta \sin (\gamma - \alpha ) + \cos \gamma \sin (\alpha - \beta ) = $

The value of $\frac{\tan 70^\circ - \tan 20^\circ}{\tan 50^\circ} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module