Considering only the principal values of inve | vedclass

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(C) Given the equation $	an^{-1}(2x) + 	an^{-1}(3x) = \frac{\pi}{4}$.Using the formula $	an^{-1}(A) + 	an^{-1}(B) = 	an^{-1}\left(\frac{A+B}{1-AB

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is a singleton

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(C) Given the equation $	an^{-1}(2x) + 	an^{-1}(3x) = \frac{\pi}{4}$.Using the formula $	an^{-1}(A) + 	an^{-1}(B) = 	an^{-1}\left(\frac{A+B}{1-AB}ight)$,we have:$	an^{-1}\left(\frac{2x+3x}{1-(2x)(3x)}ight) = \frac{\pi}{4}$.Taking tangent on both sides:$\frac{5x}{1-6x^2} = 	an\left(\frac{\pi}{4}ight) = 1$.This simplifies to $5x = 1 - 6x^2$,or $6x^2 + 5x - 1 = 0$.Factoring the quadratic equation:$(6x - 1)(x + 1) = 0$.This gives $x = \frac{1}{6}$ or $x = -1$.Since the set $A$ is defined for $x \geq 0$,we reject $x = -1$.Thus,$x = \frac{1}{6}$ is the only solution.Therefore,the set $A = \{\frac{1}{6}\}$ is a singleton set.

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

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