Q96 / 155 · WBJEE 2024 · Mathematics
If $\alpha_1, \alpha_2, \cdots, \alpha_n$ are in A.P. with common difference $\theta$, then the sum of the series $\sec \alpha_1 \sec \alpha_2+\sec \alpha_2 \sec \alpha_3+\cdots \cdot+\sec \alpha_{n-1} \sec \alpha_n=k\left(\tan \alpha_n-\tan \alpha_1\right)$ where $k=$
- A$\sin \theta$
- B$\cos \theta$
- C$\sec \theta$
- Correct answerD$\operatorname{cosec} \theta$
Solution
Hint: $\alpha_2-\alpha_1=\alpha_3-\alpha_2=\cdots=\theta$
$\begin{aligned}
& \mathrm{s}=\frac{1}{\sin \theta}\left[\frac{\sin \left(\alpha_2-\alpha_1\right)}{\cos \alpha_2 \cos \alpha_1}+\cdots+\frac{\sin \left(\alpha_n-\alpha_{n-1}\right)}{\cos \alpha_n \cos \alpha_{n-1}}\right] \\
& =\operatorname{cosec} \theta\left[\tan \alpha_n-\tan \alpha_1\right]
\end{aligned}$
Hence, the correct answer is option: 'D'.