If $a_{1},a_{2},\ldots a_{n}$ are in AP with common difference $d
eq 0$, then $\left( \sin d ight)\lbrack\sec a_{1}\sec a_{2} + \sec a_{2}\sec a_{3} + ... + \sec a_{n - 1}\sec a_{n} brack$ is equal to

Question

If $a_{1},a_{2},\ldots a_{n}$ are in AP with common difference $d 
eq 0$, then $\left( \sin d ight)\lbrack\sec a_{1}\sec a_{2} + \sec a_{2}\sec a_{3} + ... + \sec a_{n - 1}\sec a_{n}brack$ is equal to

Dr. Abubacker · Accepted Answer

(c)
Given, $d = a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = ... = a_{n} - a_{n - 1}$
$	herefore\ \ \ \left( \sin d ight)\left\lbrack \sec a_{1}\sec a_{2} + \sec a_{2}\sec a_{3} + ... + \sec a_{n - 1}\sec a_{n} ightbrack$
$= \frac{\sin d}{\cos a_{1}\cos a_{2}} + \frac{\sin d}{\cos a_{2}\cos a_{3}} + ... + \frac{\sin d}{\cos a_{n - 1}\cos a_{n}}$
$= \frac{\sin{(a_{2} - a_{1})}}{\cos a_{1}\cos a_{2}} + \frac{\sin{(a_{3} - a_{2})}}{\cos a_{2}\cos a_{3}} + ... + \frac{\sin{(a_{n} - a_{n - 1})}}{\cos{a_{n - 1}\cos a_{n}}}$
$= 	an a_{2} - 	an a_{1} + 	an a_{3} - 	an a_{2} + ... + 	an a_{n} - 	an\ a_{n - 1}$
$= 	an a_{n} - 	an a_{1}$

Q. If $a_{1},a_{2},\ldots a_{n}$ are in AP with common difference $d \neq 0$, then $\left( \sin d \right)\lbrack\sec a_{1}\sec a_{2} + \sec a_{2}\sec a_{3} + ... + \sec a_{n - 1}\sec a_{n}\rbrack$ is equal to

Explanation

Reviewed by Dr. Abubacker

Want more practice?