Solve the following equation for x: ^ -1 2 + ^ -1 3 = 4 ,0<x< 6 - Vidyadip

Question

Solve the following equation for x:
$\tan^{-1}\frac{\pi}{2}+\tan^{-1}\frac{\pi}{3}=\frac{\pi}{4},0<\text{x}<\sqrt6$

✓

Answer

Given,
$\tan^{-1}\frac{\pi}{2}+\tan^{-1}\frac{\pi}{3}=\frac{\pi}{4},0<\text{x}<\sqrt6$
$\Rightarrow\tan^{-1}\bigg[\frac{\frac{\text{x}}{2}+\frac{\text{x}}{3}}{1-\frac{\text{x}}{2}\times\frac{\text{x}}{3}}\bigg]=\frac{\pi}{4}$
$\Big\{\text{Since},\tan^{-1}\text{x}+\tan^{-1}\text{y}=\tan^{-1}\Big(\frac{\text{x}+\text{y}}{1-\text{xy}}\Big)\Big\}$
$\Rightarrow\tan^{-1}\Bigg[\frac{\frac{5\text{x}}{6}}{\frac{(6-\text{x}^2)}{16}}\Bigg]=\frac{\pi}{4}$
$\Rightarrow\tan^{-1}\Big[\frac{5\text{x}}{6-\text{x}^2}\Big]=\frac{\pi}{4}$
$\Rightarrow\frac{5\text{x}}{6-\text{x}^2}=\tan\frac{\pi}{4}$
$\Rightarrow\frac{5\text{x}}{6-\text{x}^2}=1$
$\Rightarrow5\text{x}=6\text{x}^2$
$\Rightarrow\text{x}^2+5\text{x}-6=0$
$\Rightarrow\text{x}^2+6\text{x}-\text{x}-6=0$
$\Rightarrow\text{x}(\text{x}+6)-1(\text{x}+6)=0$
$\Rightarrow(\text{x}+6)(\text{x}-1)=0$
$\Rightarrow\text{x}=-6\text{ or }\text{x}=1$
But, $0<\text{x}<\sqrt6,$ so,
$\text{x}=1$

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