If θ1, θ2, θ3, …, θn are in A.P.,

Solution:

Given $\theta_{1}, \theta_{2}, \theta_{3}, \ldots, \theta_{n}$ are in A.P., and common difference is $d$, Now we have to prove that

$\sec \theta_{1} \sec \theta_{2}+\sec \theta_{2} \sec \theta_{3}+\ldots+\sec \theta_{n-1} \sec \theta_{n}=\frac{\tan \theta_{n}-\tan \theta_{1}}{\sin d}$

On cross multiplication we get

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

We know sec $x=1 / \cos x$ using this formula we get

$\Rightarrow \frac{\sin d}{\cos \theta_{1} \cos \theta_{2}}+\frac{\text { sind }}{\cos \theta_{2} \cos \theta_{3}}+\cdots+\frac{\operatorname{sind}}{\cos \theta_{n-1} \cos \theta_{n}}=\tan \theta_{n}-\tan \theta_{1}$

Consider LHS

$\Rightarrow \mathrm{LHS}=\frac{\text { sind }}{\cos \theta_{1} \cos \theta_{2}}+\frac{\text { sind }}{\cos \theta_{2} \cos \theta_{3}}+\cdots+\frac{\text { sind }}{\cos \theta_{\mathrm{n}-1} \cos \theta_{\mathrm{n}}}$

Now we have to find value of $d$ in terms of $\theta$ so that further simplification can be made

As $\theta_{1}, \theta_{2}, \theta_{3}, \ldots, \theta_{n}$ are in AP having common difference as $d$

Hence

$\theta_{2}-\theta_{1}=d, \theta_{3}-\theta_{2}=d, \ldots, \theta_{n}-\theta_{n-1}=d$

Take sin on both sides

$\sin \left(\theta_{2}-\theta_{1}\right)=\sin d, \sin \left(\theta_{3}-\theta_{2}\right)=\sin d, \ldots, \sin \left(\theta_{n}-\theta_{n-1}\right)=\sin d$

Substitute appropriate value of sin d for each term in LHS

$\Rightarrow \mathrm{LHS}=\frac{\sin \left(\theta_{2}-\theta_{1}\right)}{\cos \theta_{1} \cos \theta_{2}}+\frac{\sin \left(\theta_{3}-\theta_{2}\right)}{\cos \theta_{2} \cos \theta_{3}}+\cdots+\frac{\sin \left(\theta_{\mathrm{n}}-\theta_{\mathrm{n}-1}\right)}{\cos \theta_{\mathrm{n}-1} \cos \theta_{\mathrm{n}}}$

We know that $\sin (a-b)=\sin a \cos b-\cos a \sin b$

Using this formula we get

$\Rightarrow \mathrm{LHS}=\frac{\sin \theta_{2} \cos \theta_{1}-\cos \theta_{2} \sin \theta_{1}}{\cos \theta_{1} \cos \theta_{2}}+\frac{\sin \theta_{3} \cos \theta_{2}-\cos \theta_{3} \sin \theta_{2}}{\cos \theta_{2} \cos \theta_{3}}+\cdots$

On simplifying we get

$=\frac{\sin \theta_{2} \cos \theta_{1}}{\cos \theta_{1} \cos \theta_{2}}-\frac{\cos \theta_{2} \sin \theta_{1}}{\cos \theta_{1} \cos \theta_{2}}+\frac{\sin \theta_{3} \cos \theta_{2}}{\cos \theta_{2} \cos \theta_{3}}-\frac{\cos \theta_{3} \sin \theta_{2}}{\cos \theta_{2} \cos \theta_{3}}+\cdots+\frac{\sin \theta_{n} \cos \theta_{n-1}}{\cos \theta_{n-1} \cos \theta_{n}}$ $-\frac{\cos \theta_{n} \sin \theta_{n-1}}{\cos \theta_{n-1} \cos \theta_{n}}$

We know that $\sin x / \cos x=\tan x$

$=\tan \theta_{2}-\tan \theta_{1}+\tan \theta_{3}-\tan \theta_{2}+\ldots+\tan \theta_{n}-\tan \theta_{n-1}$

$=-\tan \theta_{1}+\tan \theta_{n}$

$=\tan \theta_{n}-\tan \theta_{1}$

$\Rightarrow \mathrm{LHS}=\mathrm{RHS}$

Hence proved