Suppose theta and phi neq 0 are such that sec | vedclass

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(A) Since $\sec(	heta - \phi)$,$\sec 	heta$,and $\sec(	heta + \phi)$ are in $A.P.$,we have:$2 \sec 	heta = \sec(	heta - \phi) + \sec(	heta + \ph

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$\pm \sqrt{2}$

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(A) Since $\sec(	heta - \phi)$,$\sec 	heta$,and $\sec(	heta + \phi)$ are in $A.P.$,we have:$2 \sec 	heta = \sec(	heta - \phi) + \sec(	heta + \phi)$$\frac{2}{\cos 	heta} = \frac{\cos(	heta + \phi) + \cos(	heta - \phi)}{\cos(	heta - \phi) \cos(	heta + \phi)}$Using the identity $\cos(A+B) + \cos(A-B) = 2 \cos A \cos B$ and $\cos(A-B) \cos(A+B) = \cos^2 A - \sin^2 B$:$\frac{2}{\cos 	heta} = \frac{2 \cos 	heta \cos \phi}{\cos^2 	heta - \sin^2 \phi}$$\cos^2 	heta - \sin^2 \phi = \cos^2 	heta \cos \phi$$\cos^2 	heta (1 - \cos \phi) = \sin^2 \phi$$\cos^2 	heta (1 - \cos \phi) = 1 - \cos^2 \phi = (1 - \cos \phi)(1 + \cos \phi)$Since $\phi 
eq 0$,$1 - \cos \phi 
eq 0$,so:$\cos^2 	heta = 1 + \cos \phi = 2 \cos^2(\frac{\phi}{2})$$\cos 	heta = \pm \sqrt{2} \cos(\frac{\phi}{2})$Given $\cos 	heta = k \cos(\frac{\phi}{2})$,we find $k = \pm \sqrt{2}$.

Suppose $\theta$ and $\phi \neq 0$ are such that $\sec(\theta + \phi)$,$\sec \theta$,and $\sec(\theta - \phi)$ are in $A.P.$ If $\cos \theta = k \cos(\frac{\phi}{2})$ for some $k$,then $k$ is equal to

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