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$\theta$ and $\alpha$ lie in $Q_3$. If $\cos (\theta-\alpha), \cos \theta, \cos (\theta+\alpha)$ are in harmonic progression, then $\cos \theta \sec \frac{\alpha}{2}=$
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The correct answer is:
$\sqrt{2}$
Given, $\cos (\theta-\alpha), \cos \theta, \cos (\theta+\alpha)$ are in HP
$\begin{array}{lc}\because & \frac{2}{\cos \theta}=\frac{1}{\cos (\theta-\alpha)}+\frac{1}{\cos (\theta+\alpha)} \\ \Rightarrow & \frac{2}{\cos \theta}=\frac{\cos (\theta+\alpha)+\cos (\theta-\alpha)}{\cos (\theta-\alpha) \cos (\theta+\alpha)} \\ \Rightarrow & \frac{2}{\cos \theta}=\frac{2 \cos \theta \cos \alpha}{\cos ^2 \theta-\sin ^2 \alpha} \\ \Rightarrow & \cos ^2 \theta-\sin ^2 \alpha=\cos ^2 \theta \cos \alpha \\ \Rightarrow & \cos ^2 \theta(1-\cos \alpha)=\sin ^2 \alpha \\ \Rightarrow & \cos ^2 \theta=\frac{1-\cos 2 \alpha}{1-\cos ^2 \alpha}=1+\cos \alpha \\ \Rightarrow & \cos ^2 \theta=2 \cos ^2 \frac{\alpha}{2} \Rightarrow \cos \theta \sec \alpha / 2=\sqrt{2}\end{array}$
$\begin{array}{lc}\because & \frac{2}{\cos \theta}=\frac{1}{\cos (\theta-\alpha)}+\frac{1}{\cos (\theta+\alpha)} \\ \Rightarrow & \frac{2}{\cos \theta}=\frac{\cos (\theta+\alpha)+\cos (\theta-\alpha)}{\cos (\theta-\alpha) \cos (\theta+\alpha)} \\ \Rightarrow & \frac{2}{\cos \theta}=\frac{2 \cos \theta \cos \alpha}{\cos ^2 \theta-\sin ^2 \alpha} \\ \Rightarrow & \cos ^2 \theta-\sin ^2 \alpha=\cos ^2 \theta \cos \alpha \\ \Rightarrow & \cos ^2 \theta(1-\cos \alpha)=\sin ^2 \alpha \\ \Rightarrow & \cos ^2 \theta=\frac{1-\cos 2 \alpha}{1-\cos ^2 \alpha}=1+\cos \alpha \\ \Rightarrow & \cos ^2 \theta=2 \cos ^2 \frac{\alpha}{2} \Rightarrow \cos \theta \sec \alpha / 2=\sqrt{2}\end{array}$
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