Question

\[\text { If } \theta_1 , \theta_2 , \theta_3 , . . . , \theta_n \text { are in AP, whose common difference is d, then show that }\]

\[\sec \theta_1 \sec \theta_2 + \sec \theta_2 \sec \theta_3 + . . . + \sec \theta_{n - 1} \sec \theta_n = \frac{\tan \theta_n - \tan \theta_1}{\sin d} \left[ NCERT \hspace{0.167em} EXEMPLAR \right]\]

Answer 1

\[As, \theta_1 , \theta_2 , \theta_3 , . . . , \theta_n are in AP\]

\[So, d = \theta_2 - \theta_1 = \theta_3 - \theta_2 = . . . = \theta_n - \theta_{n - 1} . . . . . \left( i \right)\]

\[\text { Now }, \]

\[\text { LHS }= \sec \theta_1 \sec \theta_2 + \sec \theta_2 \sec \theta_3 + . . . + \sec \theta_{n - 1} \sec \theta_n \]

\[ = \frac{1}{\cos \theta_1 \cos \theta_2} + \frac{1}{\cos \theta_2 \cos \theta_3} + . . . + \frac{1}{\cos \theta_{n - 1} \cos \theta_n}\]

\[ = \frac{1}{\sin d}\left[ \frac{\sin d}{\cos \theta_1 \cos \theta_2} + \frac{\sin d}{\cos \theta_2 \cos \theta_3} + . . . + \frac{\sin d}{\cos \theta_{n - 1} \cos \theta_n} \right]\]

\[ = \frac{1}{\sin d}\left[ \frac{\sin\left( \theta_2 - \theta_1 \right)}{\cos \theta_1 \cos \theta_2} + \frac{\sin\left( \theta_3 - \theta_2 \right)}{\cos \theta_2 \cos \theta_3} + . . . + \frac{\sin\left( \theta_n - \theta_{n - 1} \right)}{\cos \theta_{n - 1} \cos \theta_n} \right] \left[ \text { Using } \left( i \right) \right]\]

\[ = \frac{1}{\sin d}\left[ \frac{\sin \theta_2 \cos \theta_1 - \cos \theta_2 \sin \theta_1}{cos \theta_1 cos \theta_2} + \frac{\sin \theta_3 \cos \theta_2 - \cos \theta_3 \sin \theta_2}{cos \theta_2 cos \theta_3} + . . . + \frac{\sin \theta_n \cos \theta_{n - 1} - \cos \theta_n \sin \theta_{n - 1}}{cos \theta_{n - 1} cos \theta_n} \right]\]

\[ = \frac{1}{\sin d}\left[ \frac{\sin \theta_2 \cos \theta_1}{\cos \theta_1 \cos \theta_2} - \frac{\cos \theta_2 \sin \theta_1}{\cos \theta_1 \cos \theta_2} + \frac{\sin \theta_3 \cos \theta_2}{\cos \theta_2 \cos \theta_3} - \frac{\cos \theta_3 \sin \theta_2}{\cos \theta_2 \cos \theta_3} + . . . + \frac{\sin \theta_n \cos \theta_{n - 1}}{\cos \theta_{n - 1} \cos \theta_n} - \frac{\cos \theta_n \sin \theta_{n - 1}}{\cos \theta_{n - 1} \cos \theta_n} \right]\]

\[ = \frac{1}{\sin d}\left[ \tan \theta_2 - \tan \theta_1 + \tan \theta_3 - \tan \theta_2 + . . . + \tan \theta_n - \tan \theta_{n - 1} \right]\]

\[ = \frac{1}{\sin d}\left[ - \tan \theta_1 + \tan \theta_n \right]\]

\[ = \frac{\tan \theta_n - \tan \theta_1}{\sin d}\]

\[ =\text {  RHS }\]