Question

If ${\alpha }_1,{\alpha }_2,{\alpha }_3$, ________${\alpha }_n$ are in A.P. of common difference d, then prove that $\sec {\alpha }_1\sec {\alpha }_2+\sec {\alpha }_2\sec {\alpha }_3+$_________n terms$=\frac{\tan {\alpha }_{n+1}-\tan {\alpha }_1}{\sin d}$.

Answer 1

${\alpha }_2-{\alpha }_1={\alpha }_3-{\alpha }_2={\alpha }_4-{\alpha }_3...=d$.
$T_1=\frac{1}{\cos {\alpha }_1\cos {\alpha }_2}=\frac{\sin ({\alpha }_2-{\alpha }_1)}{\cos {\alpha }_1\cos {\alpha }_2}\cdot \frac{1}{\sin d}$
or $T_1=\frac{1}{\sin d}\left[\frac{\sin {\alpha }_2\cos {\alpha }_1-\cos {\alpha }_2\sin {\alpha }_1}{\cos {\alpha }_1\cos {\alpha }_2}\right]$
$T_1=\frac{1}{\sin d}[\tan {\alpha }_2-\tan {\alpha }_1]$
$T_2=\frac{1}{\sin d}[\tan {\alpha }_3-\tan {\alpha }_2]$
$T_n=\frac{1}{\sin d}[\tan {\alpha }_{n+1}-\tan {\alpha }_n]$
$\therefore S_n=\frac{1}{\sin d}[\tan {\alpha }_{n+1}-\tan {\alpha }_1]$.