If $ an ^{-1}(2 x)+ an ^{-1}(3 x)=\frac{\pi}{4}$, where $x0$, then $x=$

Question

If $	an ^{-1}(2 x)+	an ^{-1}(3 x)=\frac{\pi}{4}$, where $x>0$, then $x=$

Aicharya · Accepted Answer

$\begin{aligned} & 	an ^{-1}(2 x)+	an ^{-1}(3 x)=\frac{\pi}{4} \ & 	herefore 	an ^{-1}\left[\frac{2 x+3 x}{1-(2 x)(3 x)}ight]=\frac{\pi}{4} \Rightarrow 	an \frac{\pi}{4}=\frac{5 x}{1-6 x^2}=1 \ & 	herefore 6 x^2+5 x-1=0 \Rightarrow(6 x-1)(x+1)=0 \Rightarrow x=-1, \frac{1}{6} \end{aligned}$ Since $x>0$, we get $x=\frac{1}{6}$

If $\tan ^{-1}(2 x)+\tan ^{-1}(3 x)=\frac{\pi}{4}$, where $x>0$, then $x=$

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