Question

If $a_1,a_2,...,a_n,$ are in AP with common difference $d\ne 0$, then $(\sin d)[sec{a}_{1}sec{a}_2+sec{a}_{2}sec{a}_3+...+sec{a}_{(n-1)}sec{a}_n]$ is equal to

• A. $cot{a}_n+cot{a}_1$
• B. $cot{a}_n-cot{a}_1$
• C. $\tan a_n-\tan a_1$
• D. $\tan a_n-\tan a_{(n-1)}$

Answer 1

The correct option is C

$\tan a_n-\tan a_1$

Explanation for the correct option:

Find the value of $\mathbf{(}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{d}\mathbf{)}\mathbf{[}\mathit{s}\mathit{e}\mathit{c}{\mathit{a}}_{\mathbf{1}}\mathit{s}\mathit{e}\mathit{c}\mathbf{}{\mathit{a}}_{\mathbf{2}}\mathbf{+}\mathbf{}\mathit{s}\mathit{e}\mathit{c}{\mathit{a}}_{\mathbf{2}}\mathit{s}\mathit{e}\mathit{c}\mathbf{}{\mathit{a}}_{\mathbf{3}}\mathbf{+}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{}\mathbf{+}\mathbf{}\mathit{s}\mathit{e}\mathit{c}{\mathit{a}}_{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}}\mathbf{}\mathit{s}\mathit{e}\mathit{c}\mathbf{}{\mathit{a}}_{\mathbf{n}}\mathbf{]}$ :

Given $a_1,a_2,\dots ...,a_n$ are in AP.

Common difference,$d=a_2-a_1=a_3-a_2=a_n–a_{(n-1)}$

$\begin{array}{rcl}& & (\sin d)[sec{a}_{1}sec{a}_2+sec{a}_{2}sec{a}_3+\dots +sec{a}_{(n-1)}sec{a}_n]\\ & =& \frac{\sin (a_2-a_1)}{\cos a_1\cos a_2}+\frac{\sin (a_3-a_2)}{\cos a_2\cos a_3}+\dots ..+\frac{\sin (a_n-a_{(n-1)})}{\cos a_{(n-1)}\cos a_n}\\ & =& \frac{(\sin a_2\cos a_1-\sin a_1\cos a_2)}{\cos a_1\cos a_2}+\frac{(\sin a_3\cos a_2-\sin a_2\cos a_3)}{\cos a_2\cos a_3}+\dots ..+\frac{(\sin a_n\cos a_{(n-1)}-\sin a_{(n-1)}\cos a_n)}{\cos a_{(n-1)}\cos a_n}\mathbf{\left[}\mathbf{\because }\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{(}\mathbf{A}\mathbf{-}\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{A}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{B}\mathbf{-}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{A}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{B}\mathbf{\right]}\\ & =& (\tan a_2–\tan a_1)+(\tan a_3–\tan a_2)+\dots .+(\tan a_n–\tan a_{n-1})\\ & =& (\tan a_n–\tan a_1)\end{array}$

Hence, the correct option is C.