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Question: If \(\cos(\theta - \alpha),\mspace{6mu}\cos\theta\)and \(\cos(\theta + \alpha)\) are in H.P., then \...

If cos⁡(θ−α), 6mucos⁡θ\cos(\theta - \alpha),\mspace{6mu}\cos\thetaand cos⁡(θ+α)\cos(\theta + \alpha) are in H.P., then cos⁡θsec⁡α2\cos\theta\sec\frac{\alpha}{2}is equal to

A

±2\pm \sqrt{2}

B

±3\pm \sqrt{3}

C

±1/2\pm 1/\sqrt{2}

D

None of these

Answer

±2\pm \sqrt{2}

Explanation

Solution

Given cos⁡(θ−α),cos⁡θ\cos(\theta - \alpha),\cos\thetaand cos⁡(θ+α)\cos(\theta + \alpha) are in H.P.

⇒ 1cos⁡(θ−α),1cos⁡θ,1cos⁡(θ+α)\frac{1}{\cos(\theta - \alpha)},\frac{1}{\cos\theta},\frac{1}{\cos(\theta + \alpha)}will be in A.P.

Hence, 2cos⁡θ=1cos⁡(θ−α)+1cos⁡(θ+α)\frac{2}{\cos\theta} = \frac{1}{\cos(\theta - \alpha)} + \frac{1}{\cos(\theta + \alpha)}

=cos⁡(α+θ)+cos⁡(θ−α)cos⁡2θ−sin⁡2α= \frac{\cos(\alpha + \theta) + \cos(\theta - \alpha)}{\cos^{2}\theta - \sin^{2}\alpha} ⇒ 2cos⁡θ=2cos⁡θcos⁡αcos⁡2θ−sin⁡2α\frac{2}{\cos\theta} = \frac{2\cos\theta\cos\alpha}{\cos^{2}\theta - \sin^{2}\alpha}

⇒ cos⁡2θ−sin⁡2α=cos⁡2θcos⁡α\cos^{2}\theta - \sin^{2}\alpha = \cos^{2}\theta\cos\alpha

⇒ cos⁡2θ(1−cos⁡α)=sin⁡2α\cos^{2}\theta(1 - \cos\alpha) = \sin^{2}\alpha

⇒ cos⁡2θ(2sin⁡2α2)=4sin⁡2α2cos⁡2α2\cos^{2}\theta\left( 2\sin^{2}\frac{\alpha}{2} \right) = 4\sin^{2}\frac{\alpha}{2}\cos^{2}\frac{\alpha}{2}

cos⁡2θsec⁡2α2=2⇒cos⁡θsec⁡α2=±2\cos^{2}\theta\sec^{2}\frac{\alpha}{2} = 2 \Rightarrow \cos\theta\sec\frac{\alpha}{2} = \pm \sqrt{2}.