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Question:

If $\theta_{1}, \theta_{2}, \theta_{3}, \ldots, \theta_{n}$ are in AP whose common difference is $d$, show that

$\sec \theta_{1} \sec \theta_{2}+\sec \theta_{2} \sec \theta_{3}+\ldots . .+\sec \theta_{n-1} \sec \theta n=\frac{\left(\tan \theta_{n}-\tan \theta_{1}\right)}{\sin d}$

Solution:

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