Since $\sec (\theta+\alpha), \sec \theta, \sec (\theta-\alpha)$ are in A.P.
$$
\begin{aligned}
& \text { Hence sec } \theta=\frac{1}{2}[\sec (\theta+\alpha)+\sec (\theta-\alpha)] \\
& \Rightarrow \sec \theta=\frac{1}{2}\left[\frac{2 \cos \theta \cdot \cos \alpha}{\cos (\theta+\alpha) \cdot \cos (\theta-\alpha)}\right] \\
& \Rightarrow \cos (\theta+\alpha) \cdot \cos (\theta-\alpha)=\cos ^2 \theta \cdot \cos \alpha \\
& \Rightarrow \cos ^2 \theta-\sin ^2 \alpha=\cos ^2 \theta \cdot \cos \alpha \\
& \Rightarrow \cos ^2 \theta=\frac{\sin ^2 \alpha}{1-\cos \alpha}=2 \cos ^2 \frac{\alpha}{2} \\
& \Rightarrow \sin ^2 \theta=1-\cos ^2 \theta=1-2 \cos ^2 \frac{\alpha}{2} \\
& \Rightarrow \sin ^2 \theta=-\cos \alpha
\end{aligned}
$$