theta and alpha lie in Q_3. If cos (theta-alp | vedclass

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(B) Given,$\cos (	heta-\alpha), \cos 	heta, \cos (	heta+\alpha)$ are in $HP$. Since they are in $HP$,their reciprocals are in $AP$: $\frac{1}{\cos

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$\sqrt{2}$

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(B) Given,$\cos (	heta-\alpha), \cos 	heta, \cos (	heta+\alpha)$ are in $HP$. Since they are in $HP$,their reciprocals are in $AP$: $\frac{1}{\cos (	heta-\alpha)}, \frac{1}{\cos 	heta}, \frac{1}{\cos (	heta+\alpha)}$ are in $AP$. Therefore,$\frac{2}{\cos 	heta} = \frac{1}{\cos (	heta-\alpha)} + \frac{1}{\cos (	heta+\alpha)}$. $\frac{2}{\cos 	heta} = \frac{\cos (	heta+\alpha) + \cos (	heta-\alpha)}{\cos (	heta-\alpha) \cos (	heta+\alpha)}$. Using the identity $\cos (A+B) + \cos (A-B) = 2 \cos A \cos B$: $\frac{2}{\cos 	heta} = \frac{2 \cos 	heta \cos \alpha}{\cos^2 	heta - \sin^2 \alpha}$. $\cos^2 	heta - \sin^2 \alpha = \cos^2 	heta \cos \alpha$. $\cos^2 	heta (1 - \cos \alpha) = \sin^2 \alpha$. $\cos^2 	heta (1 - \cos \alpha) = 1 - \cos^2 \alpha = (1 - \cos \alpha)(1 + \cos \alpha)$. Since $	heta, \alpha \in Q_3$,$\cos 	heta 
eq 0$ and $\cos \alpha 
eq 1$,so we can divide by $(1 - \cos \alpha)$: $\cos^2 	heta = 1 + \cos \alpha = 2 \cos^2 \frac{\alpha}{2}$. Taking the square root,$\cos 	heta = \pm \sqrt{2} \cos \frac{\alpha}{2}$. Since $	heta, \alpha \in Q_3$,$\cos 	heta $\cos 	heta \sec \frac{\alpha}{2} = \sqrt{2}$.

$\theta$ and $\alpha$ lie in $Q_3$. If $\cos (\theta-\alpha), \cos \theta, \cos (\theta+\alpha)$ are in harmonic progression,then $\cos \theta \sec \frac{\alpha}{2} = $

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