If sec(theta+alpha),sectheta,and sec(theta-al | vedclass

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(D) Given that $\sec(	heta+\alpha)$,$\sec	heta$,and $\sec(	heta-\alpha)$ are in arithmetic progression $(AP)$.Therefore,$2 \sec	heta = \sec(	heta

Vedclass · Accepted Answer

$\pm \sqrt{2}$

Vedclass · Answer

(D) Given that $\sec(	heta+\alpha)$,$\sec	heta$,and $\sec(	heta-\alpha)$ are in arithmetic progression $(AP)$.Therefore,$2 \sec	heta = \sec(	heta+\alpha) + \sec(	heta-\alpha)$.Using the identity $\sec x = \frac{1}{\cos x}$,we have:$\frac{2}{\cos	heta} = \frac{1}{\cos(	heta+\alpha)} + \frac{1}{\cos(	heta-\alpha)}$.$\frac{2}{\cos	heta} = \frac{\cos(	heta-\alpha) + \cos(	heta+\alpha)}{\cos(	heta+\alpha)\cos(	heta-\alpha)}$.Using the sum-to-product formula $\cos(A-B) + \cos(A+B) = 2\cos A \cos B$:$\frac{2}{\cos	heta} = \frac{2\cos	heta \cos\alpha}{\cos^2	heta - \sin^2\alpha}$.$\frac{1}{\cos	heta} = \frac{\cos	heta \cos\alpha}{\cos^2	heta - \sin^2\alpha}$.$\cos^2	heta - \sin^2\alpha = \cos^2	heta \cos\alpha$.$\cos^2	heta(1 - \cos\alpha) = \sin^2\alpha$.$\cos^2	heta(2\sin^2\frac{\alpha}{2}) = (2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2})^2$.$\cos^2	heta(2\sin^2\frac{\alpha}{2}) = 4\sin^2\frac{\alpha}{2}\cos^2\frac{\alpha}{2}$.Assuming $\sin\frac{\alpha}{2} 
eq 0$,we get $\cos^2	heta = 2\cos^2\frac{\alpha}{2}$.Taking the square root,$\cos	heta = \pm \sqrt{2} \cos\frac{\alpha}{2}$.Therefore,$\cos	heta \cdot \sec\frac{\alpha}{2} = \pm \sqrt{2}$.

List-$A$	List-$B$
$(I)$ $\tan \left(\frac{\alpha + \beta}{2}\right) =$	$(a)$ $\frac{b}{a}$
$(II)$ $\cos (\alpha + \beta) =$	$(b)$ $\frac{2ab}{a^2 + b^2}$
$(III)$ $\sin (\alpha + \beta) =$	$(c)$ $\frac{2ab}{a^2 - b^2}$
$(IV)$ $\tan (\alpha + \beta) =$	$(d)$ $\frac{a^2 - b^2}{a^2 + b^2}$

If $\sec(\theta+\alpha)$,$\sec\theta$,and $\sec(\theta-\alpha)$ are in arithmetic progression,then the value of $\cos\theta \cdot \sec\frac{\alpha}{2}$ is:

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