If tan^{-1} (ax) + tan^{-1} (3x) = frac{pi}{4 | vedclass

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

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$2$

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(A) Given the equation: $	an^{-1} (ax) + 	an^{-1} (3x) = \frac{\pi}{4}$.Using the formula $	an^{-1} A + 	an^{-1} B = 	an^{-1} \left( \frac{A+B}{1-AB} ight)$, we get:$	an^{-1} \left( \frac{ax + 3x}{1 - 3ax^2} ight) = \frac{\pi}{4}$.Taking $	an$ on both sides: $\frac{x(a+3)}{1-3ax^2} = 	an \left( \frac{\pi}{4} ight) = 1$.Substitute $x = \frac{1}{6}$ into the equation:$\frac{\frac{1}{6}(a+3)}{1 - 3a(\frac{1}{6})^2} = 1$.$\frac{\frac{a+3}{6}}{1 - \frac{3a}{36}} = 1 \implies \frac{a+3}{6} = 1 - \frac{a}{12}$.Multiply by $12$: $2(a+3) = 12 - a$.$2a + 6 = 12 - a \implies 3a = 6 \implies a = 2$.

If $\tan^{-1} (ax) + \tan^{-1} (3x) = \frac{\pi}{4}$, where $3ax^2 < 1$, then the value of $a$ for $x = \frac{1}{6}$ is...

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