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INVERSE TRIGNOMETRIC FUNCTIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINVERSE TRIGNOMETRIC FUNCTIONS5 Marks
Question
If $a_1, a_2, a_3, ..., a_n$ is an arithmetic progression with common difference d, then evaluate the following expression.
$\tan\bigg[\tan^{-1}\Big(\frac{\text{d}}{1+\text{a}_1\text{a}_2}\Big)+\tan^{-1}\Big(\frac{\text{d}}{1+\text{a}_2\text{a}_3}\Big)+\tan^{-1}\Big(\frac{\text{d}}{1+\text{a}_3\text{a}_4}\Big)+\ .....\ +\tan^{-1}\Big(\frac{\text{d}}{1+\text{a}_{\text{n}-1}\text{a}_\text{n}}\Big)\bigg]$
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Answer

We have, $a_1 = a, a_2 = a + d, a_3 = a + 2d ....$
And $d = a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = ..... = a_n - a_{n-1}$
Given that, $\tan\bigg[\tan^{-1}\Big(\frac{\text{d}}{1+\text{a}_1\text{a}_2}\Big)+\tan^{-1}\Big(\frac{\text{d}}{1+\text{a}_2\text{a}_3}\Big)+\tan^{-1}\Big(\frac{\text{d}}{1+\text{a}_3\text{a}_4}\Big)+\ .....\ +\tan^{-1}\Big(\frac{\text{d}}{1+\text{a}_{\text{n}-1}\text{a}_\text{n}}\Big)\bigg]$
$=\tan\bigg[\tan^{-1}\frac{\text{a}_2-\text{a}_1}{1+\text{a}_2.\text{a}_1}+\tan^{-1}\frac{\text{a}_3-\text{a}_2}{1+\text{a}_3.\text{a}_2}+\ ....\ +\tan^{-1}\frac{\text{a}_\text{n}-\text{a}_{\text{n}-1}}{1+\text{a}_\text{n}.\text{a}_{\text{n}-1}}\bigg]$
$=\tan\bigg[\Big(\tan^{-1}\text{a}_2-\tan^{-1}\text{a}_1\Big)+\Big(\tan^{-1}\text{a}_3-\tan^{-1}\text{a}_2\Big)+\ ....\ +\Big(\tan^{-1}\text{a}_{\text{n}}-\tan^{-1}\text{a}_{\text{n}-1}\Big)\bigg]$
$=\tan\Big[\tan^{-1}\text{a}_\text{n}-\tan^{-1}\text{a}_1\Big]$
$=\tan\Big[\tan^{-1}\frac{\text{a}_\text{n}-\text{a}_1}{1+\text{a}_\text{n}.\text{a}_1}\Big]$
$\bigg[\because\ \tan^{-1}\text{x}-\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}-\text{y}}{1+\text{xy}}\Big)\bigg]$
$=\frac{\text{a}_\text{n}-\text{a}_1}{1+\text{a}_\text{n}.\text{a}_1}\ \Big[\because\ \tan\big(\tan^{-1}\text{x}\big)=\text{x}\Big]$

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