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(B) Given the equation: $	an^{-1} x + 	an^{-1} 6x = \frac{\pi}{4}$ Using the formula $	an^{-1}(u) + 	an^{-1}(v) = 	an^{-1}\left(\frac{u+v}{1-uv}\

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is a singleton set.

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(B) Given the equation: $	an^{-1} x + 	an^{-1} 6x = \frac{\pi}{4}$ Using the formula $	an^{-1}(u) + 	an^{-1}(v) = 	an^{-1}\left(\frac{u+v}{1-uv}ight)$,we get: $	an^{-1}\left(\frac{x + 6x}{1 - (x)(6x)}ight) = \frac{\pi}{4}$ $	an^{-1}\left(\frac{7x}{1 - 6x^2}ight) = \frac{\pi}{4}$ Taking $	an$ on both sides: $\frac{7x}{1 - 6x^2} = 	an\left(\frac{\pi}{4}ight) = 1$ $7x = 1 - 6x^2$ $6x^2 + 7x - 1 = 0$ Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-7 \pm \sqrt{49 - 4(6)(-1)}}{12} = \frac{-7 \pm \sqrt{49 + 24}}{12} = \frac{-7 \pm \sqrt{73}}{12}$ Since the condition is $x \geq 0$,we reject the negative root: $x = \frac{-7 + \sqrt{73}}{12}$ Since there is only one valid value for $x$,the set $A$ is a singleton set.

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

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