If `theta_1,theta_2,theta_3, ,theta_n` are in AP, whose common difference is `d` , show that `sectheta_1s e ctheta_2+s e ctheta_2sectheta_3++sectheta_(n-1)sectheta_n=(tantheta_n-tantheta_1)/(sind)`

Explanation

To prove that

\[ \sec \theta_1 \sec \theta_2 + \sec \theta_2 \sec \theta_3 + \ldots + \sec \theta_{n-1} \sec \theta_n = \frac{\tan \theta_n - \tan \theta_1}{\sin d} \]

where \(\theta_1, \theta_2, \ldots, \theta_n\) are in arithmetic progression (AP) with common difference \(d\), we can follow these steps:

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