In a triangle ABC, tan A and tan B are the ro | vedclass

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(A) The given equation is $pq(x^{2}+1) = r^{2}x$, which can be rewritten as $x^{2} - \frac{r^{2}}{pq}x + 1 = 0$.Since $	an A$ and $	an B$ are the ro

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a right-angled triangle

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(A) The given equation is $pq(x^{2}+1) = r^{2}x$, which can be rewritten as $x^{2} - \frac{r^{2}}{pq}x + 1 = 0$.Since $	an A$ and $	an B$ are the roots, we have $	an A + 	an B = \frac{r^{2}}{pq}$ and $	an A 	an B = 1$.We know that in a $	riangle ABC$, $A + B + C = 180^{\circ}$, so $A + B = 180^{\circ} - C$.Taking tangent on both sides, $	an(A + B) = 	an(180^{\circ} - C) = -	an C$.Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$, we get $\frac{\frac{r^{2}}{pq}}{1 - 1} = -	an C$.This implies $\frac{r^{2}/pq}{0} = -	an C$, which means $	an C$ is undefined.Therefore, $C = 90^{\circ}$, which implies $	riangle ABC$ is a right-angled triangle.

In a $\triangle ABC$, $\tan A$ and $\tan B$ are the roots of the equation $pq(x^{2}+1) = r^{2}x$. Then, $\triangle ABC$ is:

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