Solve the following
Question:

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

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

Solution:

As, $\theta_{1}, \theta_{2}, \theta_{3}, \ldots, \theta_{n}$ are in AP

So, $d=\theta_{2}-\theta_{1}=\theta_{3}-\theta_{2}=\ldots=\theta_{n}-\theta_{n-1} \quad \ldots$ (i)

Now,

$\mathrm{LHS}=\sec \theta_{1} \sec \theta_{2}+\sec \theta_{2} \sec \theta_{3}+\ldots+\sec \theta_{n-1} \sec \theta_{n}$

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

$=\frac{1}{\sin d}\left[\frac{\sin d}{\cos \theta_{1} \cos \theta_{2}}+\frac{\sin d}{\cos \theta_{2} \cos \theta_{3}}+\ldots+\frac{\sin d}{\cos \theta_{n-1} \cos \theta_{n}}\right]$

$=\frac{1}{\sin d}\left[\frac{\sin \left(\theta_{2}-\theta_{1}\right)}{\cos \theta_{1} \cos \theta_{2}}+\frac{\sin \left(\theta_{3}-\theta_{2}\right)}{\cos \theta_{2} \cos \theta_{3}}+\ldots+\frac{\sin \left(\theta_{n}-\theta_{n-1}\right)}{\cos \theta_{n-1} \cos \theta_{\mathrm{n}}}\right] \quad[\mathrm{U} \sin \mathrm{g} \quad(\mathrm{i})]$

$=\frac{1}{\sin d}\left[\frac{\sin \theta_{2} \cos \theta_{1}-\cos \theta_{2} \sin \theta_{1}}{\cos \theta_{1} \cos \theta_{2}}+\frac{\sin \theta_{3} \cos \theta_{2}-\cos \theta_{3} \sin \theta_{2}}{\cos \theta_{2} \cos \theta_{3}}+\ldots+\frac{\sin \theta_{n} \cos \theta_{n-1}-\cos \theta_{n} \sin \theta_{n-1}}{\cos \theta_{n-1} \cos \theta_{n}}\right]$

$=\frac{1}{\sin d}\left[\frac{\sin \theta_{2} \cos \theta_{1}}{\cos \theta_{1} \cos \theta_{2}}-\frac{\cos \theta_{2} \sin \theta_{1}}{\cos \theta_{1} \cos \theta_{2}}+\frac{\sin \theta_{3} \cos \theta_{2}}{\cos \theta_{2} \cos \theta_{3}}-\frac{\cos \theta_{3} \sin \theta_{2}}{\cos \theta_{2} \cos \theta_{3}}+\ldots+\frac{\sin \theta_{n} \cos \theta_{n-1}}{\cos \theta_{n-1} \cos \theta_{\mathrm{n}}}-\frac{\cos \theta_{n} \sin \theta_{n-1}}{\cos \theta_{n-1} \cos \theta_{n}}\right]$

$=\frac{1}{\sin d}\left[\tan \theta_{2}-\tan \theta_{1}+\tan \theta_{3}-\tan \theta_{2}+\ldots+\tan \theta_{n}-\tan \theta_{n-1}\right]$

$=\frac{1}{\sin d}\left[-\tan \theta_{1}+\tan \theta_{n}\right]$

$=\frac{\tan \theta_{n}-\tan \theta_{1}}{\sin d}$

= RHS

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