If $\sec ( heta+\alpha), \sec heta$ and $\sec ( heta-\alpha)$ are arithmetic progression, then $\sin ^2 heta=$

Question

If $\sec (	heta+\alpha), \sec 	heta$ and $\sec (	heta-\alpha)$ are arithmetic progression, then $\sin ^2 	heta=$

Aicharya · Accepted Answer

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

If $\sec (\theta+\alpha), \sec \theta$ and $\sec (\theta-\alpha)$ are arithmetic progression, then $\sin ^2…

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