If sec (theta+alpha), sec theta and sec (thet | vedclass

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(D) Since $\sec (	heta+\alpha), \sec 	heta, \sec (	heta-\alpha)$ are in $A$.$P$.,we have: $2 \sec 	heta = \sec (	heta+\alpha) + \sec (	heta-\alp

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$-\cos \alpha$

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(D) Since $\sec (	heta+\alpha), \sec 	heta, \sec (	heta-\alpha)$ are in $A$.$P$.,we have: $2 \sec 	heta = \sec (	heta+\alpha) + \sec (	heta-\alpha)$ $\Rightarrow \frac{2}{\cos 	heta} = \frac{1}{\cos (	heta+\alpha)} + \frac{1}{\cos (	heta-\alpha)}$ $\Rightarrow \frac{2}{\cos 	heta} = \frac{\cos (	heta-\alpha) + \cos (	heta+\alpha)}{\cos (	heta+\alpha) \cos (	heta-\alpha)}$ Using $\cos (A+B) + \cos (A-B) = 2 \cos A \cos B$: $\Rightarrow \frac{2}{\cos 	heta} = \frac{2 \cos 	heta \cos \alpha}{\cos^2 	heta - \sin^2 \alpha}$ $\Rightarrow \cos^2 	heta - \sin^2 \alpha = \cos^2 	heta \cos \alpha$ $\Rightarrow \cos^2 	heta (1 - \cos \alpha) = \sin^2 \alpha$ $\Rightarrow \cos^2 	heta (1 - \cos \alpha) = 1 - \cos^2 \alpha = (1 - \cos \alpha)(1 + \cos \alpha)$ Since $1 - \cos \alpha 
eq 0$ (assuming $\alpha 
eq 2n\pi$),we have: $\cos^2 	heta = 1 + \cos \alpha$ $\Rightarrow \sin^2 	heta = 1 - \cos^2 	heta = 1 - (1 + \cos \alpha) = -\cos \alpha$

If $\sec (\theta+\alpha), \sec \theta$ and $\sec (\theta-\alpha)$ are in arithmetic progression,then $\sin ^2 \theta=$

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