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

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(A) Given that $\cos (	heta - \alpha ), \cos 	heta, \cos (	heta + \alpha )$ are in $H.P.$Therefore, their reciprocals $\frac{1}{\cos (	heta - \alp

Vedclass · Accepted Answer

$\pm \sqrt{2}$

Vedclass · Answer

(A) Given that $\cos (	heta - \alpha ), \cos 	heta, \cos (	heta + \alpha )$ are in $H.P.$Therefore, their reciprocals $\frac{1}{\cos (	heta - \alpha )}, \frac{1}{\cos 	heta}, \frac{1}{\cos (	heta + \alpha )}$ are in $A.P.$This implies $\frac{2}{\cos 	heta} = \frac{1}{\cos (	heta - \alpha )} + \frac{1}{\cos (	heta + \alpha )}$.Using the formula $\cos (A+B) + \cos (A-B) = 2 \cos A \cos B$, we get $\frac{2}{\cos 	heta} = \frac{\cos (	heta + \alpha ) + \cos (	heta - \alpha )}{\cos (	heta - \alpha ) \cos (	heta + \alpha )} = \frac{2 \cos 	heta \cos \alpha}{\cos^2 	heta - \sin^2 \alpha}$.Simplifying, $\frac{1}{\cos 	heta} = \frac{\cos 	heta \cos \alpha}{\cos^2 	heta - \sin^2 \alpha}$.$\cos^2 	heta - \sin^2 \alpha = \cos^2 	heta \cos \alpha$.$\cos^2 	heta (1 - \cos \alpha) = \sin^2 \alpha$.Using $1 - \cos \alpha = 2 \sin^2 \frac{\alpha}{2}$ and $\sin^2 \alpha = 4 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2}$, we have $\cos^2 	heta (2 \sin^2 \frac{\alpha}{2}) = 4 \sin^2 \frac{\alpha}{2} \cos^2 \frac{\alpha}{2}$.$\cos^2 	heta = 2 \cos^2 \frac{\alpha}{2}$.$\cos^2 	heta \sec^2 \frac{\alpha}{2} = 2$.Taking the square root, $\cos 	heta \sec \frac{\alpha}{2} = \pm \sqrt{2}$.

If $\cos (\theta - \alpha ), \cos \theta$ and $\cos (\theta + \alpha )$ are in $H.P.$, then $\cos \theta \sec \frac{\alpha }{2}$ is equal to:

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