If 3 tan^{-1} x + cot^{-1} x = pi,then x is e | vedclass

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(B) Given the equation: $3 	an^{-1} x + \cot^{-1} x = \pi$. We know that $	an^{-1} x + \cot^{-1} x = \frac{\pi}{2}$. We can rewrite the given equati

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(B) Given the equation: $3 	an^{-1} x + \cot^{-1} x = \pi$. We know that $	an^{-1} x + \cot^{-1} x = \frac{\pi}{2}$. We can rewrite the given equation as: $2 	an^{-1} x + (	an^{-1} x + \cot^{-1} x) = \pi$. Substituting the identity $	an^{-1} x + \cot^{-1} x = \frac{\pi}{2}$ into the equation: $2 	an^{-1} x + \frac{\pi}{2} = \pi$. Subtracting $\frac{\pi}{2}$ from both sides: $2 	an^{-1} x = \frac{\pi}{2}$. Dividing by $2$: $	an^{-1} x = \frac{\pi}{4}$. Taking the tangent of both sides: $x = 	an(\frac{\pi}{4}) = 1$.

If $3 \tan^{-1} x + \cot^{-1} x = \pi$,then $x$ is equal to:

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