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

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

Aicharya · Accepted Answer

$\begin{aligned} & 	an ^{-1}(2 x)+	an ^{-1}(3 x)=\frac{\pi}{4} \ & \Rightarrow 	an ^{-1}\left[\frac{2 x+3 x}{1-(2 x)(3 x)}ight]=\frac{\pi}{4} \ & \Rightarrow \frac{5 x}{1-6 x^2}=	an \frac{\pi}{4}=1 \ & \Rightarrow 6 x^2+5 x-1=0 \ & \Rightarrow(x+1)(6 x-1)=0 \ & \Rightarrow x=-1 	ext { or } x=\frac{1}{6} \ & 	ext {But, } x \geq 0 \ & 	herefore \quad x=\frac{1}{6} \ & 	herefore \quad \mathrm{~A} 	ext { is a singleton set. } \end{aligned}$

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

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