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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:
$\tan^{-1}\Big(\frac{\text{x}-2}{\text{x}-4}\Big)+\tan^{-1}\Big(\frac{\text{x}+2}{\text{x}+4}\Big)=\frac{\pi}{4}$
✓

Answer

$\tan^{-1}\Big(\frac{\text{x}-2}{\text{x}-4}\Big)+\tan^{-1}\Big(\frac{\text{x}+2}{\text{x}+4}\Big)=\frac{\pi}{4}$
We know
$\because\ \tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)$
$\Rightarrow\tan^{-1}\Bigg(\frac{\frac{\text{x}-2}{\text{x}-4}+\frac{\text{x}+2}{\text{x}+4}}{1-\frac{\text{x}-2}{\text{x}-4}\times\frac{\text{x}+2}{\text{x}+4}}\Bigg)=\frac{\pi}4{}$
$\Rightarrow\tan^{-1}\Bigg(\frac{\frac{\text{x}^2+2\text{x}-8+\text{x}^2-2\text{x}-8}{(\text{x}-4)(\text{x}+4)}}{\frac{\text{x}^2-16-\text{x}^2+4}{(\text{x}-4)(\text{x}+4)}}\Bigg)=\frac{\pi}{4}$
$\Rightarrow\frac{2\text{x}^2-16}{-12}=\tan\frac{\pi}{4}$
$\Rightarrow\frac{2\text{x}^2-16}{-12}=1$
$\Rightarrow2\text{x}^2-16=-12$
$\Rightarrow2\text{x}^2=4$
$\Rightarrow\text{x}^2=2$
$\Rightarrow\text{x}=\pm\sqrt2$

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