If α1, α2, …, αn are in A.P. with common difference θ, then the sum of the series sec α1 sec α2 + sec α2 sec α3 + …. + sec αn − 1 sec αn = k(tan αn − tan α1) where k =
cosec θ
A sum of products of secants of angles in A.P. telescopes once each term is written as a difference of tangents. The basic identity is \(\tan B-\tan A=\dfrac{\sin(B-A)}{\cos A\cos B}\).
Step 1 — For consecutive terms, \(\alpha_{r+1}-\alpha_{r}=\theta\), so \(\tan\alpha_{r+1}-\tan\alpha_{r}=\dfrac{\sin\theta}{\cos\alpha_{r}\cos\alpha_{r+1}}\).
Step 2 — Hence \(\sec\alpha_{r}\sec\alpha_{r+1}=\dfrac{1}{\sin\theta}\left(\tan\alpha_{r+1}-\tan\alpha_{r}\right)\).
Step 3 — Summing from r = 1 to n − 1, the inner tangents cancel: \(\sum_{r=1}^{n-1}\sec\alpha_{r}\sec\alpha_{r+1}=\dfrac{1}{\sin\theta}\left(\tan\alpha_{n}-\tan\alpha_{1}\right)\).
Step 4 — Comparing with \(k(\tan\alpha_{n}-\tan\alpha_{1})\) gives \(k=\dfrac{1}{\sin\theta}=\operatorname{cosec}\theta\).
The factor is the reciprocal of sin θ, so sin θ itself, cos θ and sec θ do not make the identity hold (for example with θ small, the left side is large while sin θ times the tangent difference is tiny).
cosec θ option is correct.
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