If $\theta_{1}, \theta_{2}, \theta_{3}, \ldots, \theta_{n}$ are in AP whose common difference is $d$, show that
$\sec \theta_{1} \sec \theta_{2}+\sec \theta_{2} \sec \theta_{3}+\ldots . .+\sec \theta_{n-1} \sec \theta n=\frac{\left(\tan \theta_{n}-\tan \theta_{1}\right)}{\sin d}$
Show that: $\sec \theta_{1} \sec \theta_{2}+\sec \theta_{2} \sec \theta_{3}+\ldots .+\sec \theta_{n-1} \sec \theta n=\frac{\left(\tan \theta_{n}-\tan \theta_{1}\right)}{\sin d}$
Given: Given AP is $\theta_{1}, \theta_{2}, \theta_{3}, \ldots ., \theta_{n}$
$a=\theta_{1}, a_{2}=\theta_{2}$ and $d=\theta_{2}-\theta_{1}=\theta_{3}-\theta_{2}=\theta_{4}-\theta_{3}=\ldots \ldots \ldots \ldots=\theta_{n}-\theta_{n-1}$
$\sec \theta_{1} \sec \theta_{2}+\sec \theta_{2} \sec \theta_{3}+\ldots .+\sec \theta_{n-1} \sec \theta n=\frac{1}{\cos \theta 1} \times \frac{1}{\cos \theta 2}+\frac{1}{\cos \theta 2} \times \frac{1}{\cos \theta 3}$
$+\ldots \ldots \ldots+\frac{1}{\cos \theta n-1} \times \frac{1}{\cos \theta n}$
Multiply both side by sin d
[NOTE: $\sin (x-y)=\sin x \cos y-\cos x \sin y, \& \sec \theta \times \cos \theta=1]$
By using above formula on R.H.S., we get
R.H.S. $=\tan \theta_{2}-\tan \theta_{1}+\tan \theta_{3}-\tan \theta_{2}+\tan \theta_{4}-\tan \theta_{3} \ldots \ldots \ldots \ldots+\tan \theta_{n}-\tan \theta_{n-1}$
R.H.S. $=\tan \theta_{n}-\tan \theta_{1}$ (All the remaining term cancle out)
$\sin d\left(\sec \theta_{1} \sec \theta_{2}+\sec \theta_{2} \sec \theta_{3}+\ldots .+\sec \theta_{n-1} \sec \theta n\right)=\tan \theta_{n}-\tan \theta_{1}$ (Divide $\sin d$ on both sides), we get
$\sec \theta_{1} \sec \theta_{2}+\sec \theta_{2} \sec \theta_{3}+\ldots .+\sec \theta_{n-1} \sec \theta n=\frac{\left(\tan \theta_{n}-\tan \theta_{1}\right)}{\sin d}$
HENCE PROVED