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If sec $(\theta-\alpha)$, sec $\theta$ and sec $(\theta+\alpha)$ are in AP, where cos $\alpha \neq 1$, then what is the value of $\sin ^{2} \theta+\cos \alpha$ ? $[2018-\mathrm{II}]$
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$\frac{2}{\cos \theta}=\frac{\cos (\theta+\alpha)+\cos (\theta-\alpha)}{\cos (\theta+\alpha) \cos (\theta-\alpha)}=\frac{2 \cos \theta \cdot \cos \alpha}{\cos ^{2} \theta-\sin ^{2} \alpha}$
$\Rightarrow \cos ^{2} \theta \cos \alpha=\cos ^{2} \theta-\sin ^{2} \alpha$
$\Rightarrow \sin ^{2} \alpha=\cos ^{2} \theta(1-\cos \alpha)$
$\Rightarrow \cos ^{2} \theta=\frac{\sin ^{2} \alpha}{1-\cos \alpha}=1+\cos \alpha$
$\Rightarrow \quad 1-\sin ^{2} \theta=1+\cos \alpha$
$\Rightarrow \sin ^{2} \theta+\cos \alpha=0$
$\Rightarrow \cos ^{2} \theta \cos \alpha=\cos ^{2} \theta-\sin ^{2} \alpha$
$\Rightarrow \sin ^{2} \alpha=\cos ^{2} \theta(1-\cos \alpha)$
$\Rightarrow \cos ^{2} \theta=\frac{\sin ^{2} \alpha}{1-\cos \alpha}=1+\cos \alpha$
$\Rightarrow \quad 1-\sin ^{2} \theta=1+\cos \alpha$
$\Rightarrow \sin ^{2} \theta+\cos \alpha=0$
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