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

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(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 $2 \tan ^2 \theta=$

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