If cos (theta-alpha), cos theta and cos (thet | vedclass

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

Question

(A) Given that $\cos (	heta-\alpha), \cos 	heta, \cos (	heta+\alpha)$ are in harmonic progression $(HP)$.Therefore,$\frac{1}{\cos (	heta-\alpha)},

Vedclass · Accepted Answer

$	an ^2 \frac{\alpha}{2}-1$

Vedclass · Answer

(A) Given that $\cos (	heta-\alpha), \cos 	heta, \cos (	heta+\alpha)$ are in harmonic progression $(HP)$.Therefore,$\frac{1}{\cos (	heta-\alpha)}, \frac{1}{\cos 	heta}, \frac{1}{\cos (	heta+\alpha)}$ are in arithmetic progression $(AP)$.So,$\frac{2}{\cos 	heta} = \frac{1}{\cos (	heta-\alpha)} + \frac{1}{\cos (	heta+\alpha)}$.$\frac{2}{\cos 	heta} = \frac{\cos (	heta+\alpha) + \cos (	heta-\alpha)}{\cos (	heta-\alpha) \cos (	heta+\alpha)}$.Using the formula $\cos (A+B) + \cos (A-B) = 2 \cos A \cos B$,we get:$\frac{2}{\cos 	heta} = \frac{2 \cos 	heta \cos \alpha}{\cos (	heta-\alpha) \cos (	heta+\alpha)}$.$\cos^2 	heta \cos \alpha = \cos (	heta-\alpha) \cos (	heta+\alpha)$.Using $\cos (A-B) \cos (A+B) = \cos^2 A - \sin^2 B$,we get:$\cos^2 	heta \cos \alpha = \cos^2 	heta - \sin^2 \alpha$.$\sin^2 \alpha = \cos^2 	heta (1 - \cos \alpha)$.$\cos^2 	heta = \frac{\sin^2 \alpha}{1 - \cos \alpha} = \frac{4 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2}}{2 \sin^2 \frac{\alpha}{2}} = 2 \cos^2 \frac{\alpha}{2}$.Now,$	an^2 	heta = \sec^2 	heta - 1 = \frac{1}{\cos^2 	heta} - 1 = \frac{1}{2 \cos^2 \frac{\alpha}{2}} - 1 = \frac{1}{2} \sec^2 \frac{\alpha}{2} - 1$.$2 	an^2 	heta = \sec^2 \frac{\alpha}{2} - 2 = (1 + 	an^2 \frac{\alpha}{2}) - 2 = 	an^2 \frac{\alpha}{2} - 1$.

If $\cos (\theta-\alpha), \cos \theta$ and $\cos (\theta+\alpha)$ are in harmonic progression,then $2 \tan ^2 \theta=$

Similar Questions

Given that the roots of $x^3+3px^2+3qx+r=0$ are in harmonic progression,then:

If $x_1, x_3$ are the roots of $A x^2 - 4 x + 1 = 0$ and $x_2, x_4$ are the roots of $B x^2 - 6 x + 1 = 0$ such that $x_1, x_2, x_3, x_4$ are in harmonic progression,then $\frac{B+A}{B-A} = $

Which number should be added to the numbers $13, 15, 19$ so that the resulting numbers are the consecutive terms of a $H.P.$?

If $\frac{a^{n + 1} + b^{n + 1}}{a^n + b^n}$ is the harmonic mean between $a$ and $b$,then the value of $n$ is

If $\frac{1}{4}, a, b, \frac{1}{19}$ form a $H.P.$,then the values of $a$ and $b$ are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module