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 $\sin \theta_1 + \sin \theta_2 + \sin \theta_3 = 3$,then $\cos \theta_1 + \cos \theta_2 + \cos \theta_3 = $

$\frac{\sin^2 A - \sin^2 B}{\sin A \cos A - \sin B \cos B} = $

The value of $\frac{3 + \cot 76^{\circ} \cot 16^{\circ}}{\cot 76^{\circ} + \cot 16^{\circ}}$ is:

If $x = r \cos \theta \cos \phi$,$y = r \cos \theta \sin \phi$,and $z = r \sin \theta$,then the value of $x^{2} + y^{2} + z^{2}$ is

If $\pi < \alpha < \frac{3\pi}{2}$,then $\sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} + \sqrt{\frac{1 + \cos \alpha}{1 - \cos \alpha}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module