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(D) Given that $\alpha_1, \alpha_2, \cdots, \alpha_n$ are in $A$.$P$. with common difference $	heta$,so $\alpha_{i+1} - \alpha_i = 	heta$ for all $i

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$\operatorname{cosec} 	heta$

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(D) Given that $\alpha_1, \alpha_2, \cdots, \alpha_n$ are in $A$.$P$. with common difference $	heta$,so $\alpha_{i+1} - \alpha_i = 	heta$ for all $i = 1, 2, \cdots, n-1$. The general term of the series is $T_i = \sec \alpha_i \sec \alpha_{i+1} = \frac{1}{\cos \alpha_i \cos \alpha_{i+1}}$. We can write $T_i = \frac{1}{\sin 	heta} \cdot \frac{\sin(\alpha_{i+1} - \alpha_i)}{\cos \alpha_i \cos \alpha_{i+1}} = \operatorname{cosec} 	heta (	an \alpha_{i+1} - 	an \alpha_i)$. The sum of the series is $S = \sum_{i=1}^{n-1} T_i = \operatorname{cosec} 	heta \sum_{i=1}^{n-1} (	an \alpha_{i+1} - 	an \alpha_i)$. This is a telescoping sum: $S = \operatorname{cosec} 	heta [(	an \alpha_2 - 	an \alpha_1) + (	an \alpha_3 - 	an \alpha_2) + \cdots + (	an \alpha_n - 	an \alpha_{n-1})]$. Simplifying,we get $S = \operatorname{cosec} 	heta (	an \alpha_n - 	an \alpha_1)$. Comparing this with $k(	an \alpha_n - 	an \alpha_1)$,we find $k = \operatorname{cosec} 	heta$.

If $\alpha_1, \alpha_2, \cdots, \alpha_n$ are in $A$.$P$. with common difference $\theta$,then the sum of the series $\sec \alpha_1 \sec \alpha_2 + \sec \alpha_2 \sec \alpha_3 + \cdots + \sec \alpha_{n-1} \sec \alpha_n = k(\tan \alpha_n - \tan \alpha_1)$,where $k=$

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