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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions5 Marks
Question
Solve the following equation for x:
$\tan^{-1}2\text{x}+\tan^{-1}3\text{x}=\text{n}\pi+\frac{3\pi}{4}$
✓

Answer

Given$\tan^{-1}2\text{x}+\tan^{-1}3\text{x}=\text{n}\pi+\frac{3\pi}{4}....(1)$
$\Rightarrow\tan^{-1}\Big(\frac{2\text{x}+3\text{x}}{1-2\text{x}\times3\text{x}}\Big)=\text{n}\pi+\frac{3\pi}{4}$
$\Big\{\text{Since}\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x+y}}{1-\text{xy}}\Big)\text{ if }\text{xy}<1\Big\}$
$\Rightarrow\tan^{-1}\Big(\frac{5\text{x}}{1-6\text{x}^2}\Big)=\text{n}\pi+\frac{3\pi}{4},6\text{x}^2<1$
$\Rightarrow\frac{5\text{x}}{1-6\text{x}^2}=\tan\Big(\text{n}\pi+\frac{3\pi}{4}\Big),6\text{x}^2<1$
$\Rightarrow\frac{5\text{x}}{1-6\text{x}^2}-1,6\text{x}^2<1$
$\Rightarrow5\text{x}=-1+6\text{x}^2,6\text{x}^2<1$
$\Rightarrow6\text{x}^2-5\text{x}-1=0,\text{x}^2<\frac{1}{6}$
$\Rightarrow6\text{x}^2-6\text{x}+\text{x}-1=0,-\frac{1}{\sqrt6}<\text{x}<\frac{1}{6}$
$\Rightarrow6\text{x}(\text{x}-1)+1(\text{x}-1)=0,-\frac{1}{\sqrt6}<\text{x}<\frac{1}{6}$
$\Rightarrow(6\text{x}+1)(\text{x}-1)=0$
$\Rightarrow6\text{x}+1=0$ or $\text{x}-1=0$
$\Rightarrow\text{x}=-\frac{1}{6}$ or $\text{x}=1$
Since, $\text{x}=1\notin\Big(-\frac{1}{\sqrt6},\frac{1}{\sqrt6}\Big)$
So,
x = 1 is not root of the given equation (i).
Since,
$\text{x}=1\in\Big(-\frac{1}{\sqrt6},\frac{1}{\sqrt6}\Big)$
So,
$\text{x}=-\frac{1}{6}$ is the root of the given equation (i).
Hence,
$\text{x}=-\frac{1}{6}$

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