Suppose theta and phi (neq 0) are such that s | vedclass

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(A) Given that $\sec(	heta - \phi)$,$\sec	heta$,and $\sec(	heta + \phi)$ are in $A.P.$Therefore,$2 \sec	heta = \sec(	heta - \phi) + \sec(	heta +

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$\pm \sqrt{2}$

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(A) Given that $\sec(	heta - \phi)$,$\sec	heta$,and $\sec(	heta + \phi)$ are in $A.P.$Therefore,$2 \sec	heta = \sec(	heta - \phi) + \sec(	heta + \phi)$.$\frac{2}{\cos	heta} = \frac{1}{\cos(	heta - \phi)} + \frac{1}{\cos(	heta + \phi)}$$\frac{2}{\cos	heta} = \frac{\cos(	heta + \phi) + \cos(	heta - \phi)}{\cos(	heta - \phi) \cos(	heta + \phi)}$Using the identity $\cos(A+B) + \cos(A-B) = 2 \cos A \cos B$:$\frac{2}{\cos	heta} = \frac{2 \cos	heta \cos\phi}{\cos^2	heta - \sin^2\phi}$$\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 we can divide by $(1 - \cos\phi)$:$\cos^2	heta = 1 + \cos\phi = 2 \cos^2(\frac{\phi}{2})$Taking the square root on both sides:$\cos	heta = \pm \sqrt{2} \cos(\frac{\phi}{2})$Comparing this with $\cos	heta = k \cos(\frac{\phi}{2})$,we get $k = \pm \sqrt{2}$.

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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