Considering only the Principal values of inve | vedclass

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(D) Given the equation: $	an^{-1}(2x) + 	an^{-1}(3x) = \frac{\pi}{4}$ Using the identity $	an^{-1}(A) + 	an^{-1}(B) = 	an^{-1}\left(\frac{A+B}{1-

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is a singleton set.

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(D) Given the equation: $	an^{-1}(2x) + 	an^{-1}(3x) = \frac{\pi}{4}$ Using the identity $	an^{-1}(A) + 	an^{-1}(B) = 	an^{-1}\left(\frac{A+B}{1-AB}ight)$,we get: $	an^{-1}\left(\frac{2x + 3x}{1 - (2x)(3x)}ight) = \frac{\pi}{4}$ $\frac{5x}{1 - 6x^2} = 	an\left(\frac{\pi}{4}ight)$ $\frac{5x}{1 - 6x^2} = 1$ $5x = 1 - 6x^2$ $6x^2 + 5x - 1 = 0$ Factoring the quadratic equation: $6x^2 + 6x - x - 1 = 0$ $6x(x + 1) - 1(x + 1) = 0$ $(6x - 1)(x + 1) = 0$ 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 $A = \{x \geq 0 \mid \tan^{-1}(2x) + \tan^{-1}(3x) = \frac{\pi}{4}\}$

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