Considering only the principal values of an inverse function, the set $\mathrm{A}=\left\{x \geq 0 / an ^{-1} x+ an ^{-1} 6 x=\frac{\pi}{4} ight\}$

Question

Considering only the principal values of an inverse function, the set $\mathrm{A}=\left\{x \geq 0 / 	an ^{-1} x+	an ^{-1} 6 x=\frac{\pi}{4}ight\}$

Aicharya · Accepted Answer

Consider, $	an ^{-1} x+	an ^{-1} 6 x=\frac{\pi}{4}$ $\begin{aligned} & \Rightarrow 	an ^{-1}\left(\frac{7 x}{1-6 x^2}ight)=\frac{\pi}{4} \ & \quad \quad \ldots\left[	an ^{-1}(x)+	an ^{-1}(y)=	an ^{-1}\left(\frac{x+y}{1-x y}ight)ight] \ & \Rightarrow \frac{7 x}{1-6 x^2}=1 \ & \Rightarrow 7 x=1-6 x^2 \ & \Rightarrow 6 x^2+7 x-1=0 \ & \Rightarrow x=\frac{-7 \pm \sqrt{73}}{12} \ & \quad 	ext { Since } x \geq 0 \ & 	herefore \quad x=\frac{-7+\sqrt{73}}{12} \end{aligned}$ $	herefore \quad$ A is a singleton set.

Considering only the principal values of an inverse function, the set $\mathrm{A}=\left\{x \geq 0 / \tan…

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