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Inverse Trigonometric Functions — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSInverse Trigonometric Functions5 Marks
Question
Solve the following equation for x:
$\cot^{-1}\text{x}-\cot^{-1}(\text{x}+2)=\frac{\pi}{12},\text{x}>0$
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Answer

$\Rightarrow\cot^{-1}(\text{x})-\cot^{-1}(\text{x}+2)=\frac{\pi}{12}$
$\Rightarrow \tan^{-1}\Big(\frac{1}{\text{x}}\Big)+\cot^{-1}\Big(\frac{1}{\text{x}+2}\Big)=\frac{\pi}{2}$
$\Big[\because\ \cot^{-1}\text{x}=\tan^{-1}\frac{1}{\text{x}}\Big]$
$\Rightarrow\tan^{-1}\Bigg(\frac{\frac{1}{\text{x}}-\frac{1}{\text{x}+2}}{1+\frac{1}{\text{x}(\text{x}+2)}}\Bigg)=\frac{\pi}{12}$
$\Rightarrow\Bigg(\frac{\frac{2}{\text{x}(\text{x}+2)}}{\frac{\text{x}^2+2\text{x}+1}{\text{x}(\text{x}+2)}}\Bigg)=\frac{\pi}{12}$
$\Rightarrow\tan^{-1}\Big(\frac{2}{\text{x}^2+2\text{x}+1}\Big)=\frac{\pi}{12}$
$\Rightarrow\Big(\frac{2}{\text{x}^2+2\text{x}+1}\Big)=\tan\frac{\pi}{12}$
$\Rightarrow\Big(\frac{2}{\text{x}^2+2\text{x}+1}\Big)=\tan\Big(\frac{\pi}{3}-\frac{\pi}{4}\Big)$
$\Rightarrow\Big(\frac{2}{\text{x}^2+2\text{x}+1}\Big)=\frac{\tan\frac{\pi}{3}+\tan\frac{\pi}{4}}{1+\tan\frac{\pi}{3}\times\tan\frac{\pi}{4}}$
$\Rightarrow\Big(\frac{2}{\text{x}^2+2\text{x}+1}\Big)=\frac{\sqrt3-1}{\sqrt3+1}$
$\Rightarrow\Big(\frac{2}{\text{x}^2+2\text{x}+1}\Big)=\frac{\sqrt3-1}{\sqrt3+1}\times\frac{\sqrt3+1}{\sqrt3+1}$
$\Rightarrow\Big(\frac{2}{\text{x}^2+2\text{x}+1}\Big)=\frac{2}{\big(\sqrt3+1\big)^2}$
$\Rightarrow\frac{1}{(\text{x}+1)^2}\frac{1}{\big(\sqrt3+1\big)^2}$
$\Rightarrow\text{x}+1=\sqrt3+1$
$\Rightarrow\text{x}=\sqrt3$

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