Considering only the Principal values of inverse function, the set $\left\{x \geq 0 / an ^{-1}(2 x)+ an ^{-1} 3 x=\frac{\pi}{4} ight\}$

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Considering only the Principal values of inverse function, the set $\left\{x \geq 0 / 	an ^{-1}(2 x)+	an ^{-1} 3 x=\frac{\pi}{4}ight\}$

Aicharya · Accepted Answer

$\begin{aligned} & 	an ^{-1}(2 x)+	an ^{-1}(3 x)=\frac{\pi}{4} \ & \Rightarrow 	an ^{-1} \frac{2 x+3 x}{1-2 x 	imes 3 x}=	an ^{-1}(1) \ & \Rightarrow \frac{5 x}{1-6 x^2}=1 \ & \Rightarrow 5 x=1-6 x^2 \ & \Rightarrow 6 x^2+5 x-1=0 \ & \Rightarrow(x+1)(6 x-1)=0\end{aligned}$ $\Rightarrow x=-1$ or $x=\frac{1}{6}$ But $x \geq 0$ Hence $x=\frac{1}{6}$ (only) i.e., single ton set

Considering only the Principal values of inverse function, the set $\left\{x \geq 0 / \tan ^{-1}(2 x)+\tan…

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