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In a $\triangle A B C$, tan $A$ and $\tan B$ are the roota $p q\left(x^{2}+1\right)=r^{2} x$ Then, $\Delta A B C$ is
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The correct answer is:
a right angled triangle
Given equation can be rewritten as $x^{2}-\frac{r^{2}}{p q} x+1=0$
$\therefore \quad \tan A+\tan B=\frac{r^{2}}{p q}$
and $\tan A \tan B=1$
We know, $A+B+C=180^{\circ}$
$\Rightarrow \quad(A+B)=180^{\circ}-C$
$\Rightarrow \quad \tan (A+B)=\tan \left(180^{\circ}-C\right)$
$\Rightarrow \quad \frac{\tan A+\tan B}{1-\tan A \tan B}=-\tan C$
$\Rightarrow \quad \frac{r^{2} / p q}{1-1}=-\tan C$
$\Rightarrow \quad \infty=-\tan C \Rightarrow C=90^{\circ}$
Hence, $\triangle A B C$ is a right angled triangle.
$\therefore \quad \tan A+\tan B=\frac{r^{2}}{p q}$
and $\tan A \tan B=1$
We know, $A+B+C=180^{\circ}$
$\Rightarrow \quad(A+B)=180^{\circ}-C$
$\Rightarrow \quad \tan (A+B)=\tan \left(180^{\circ}-C\right)$
$\Rightarrow \quad \frac{\tan A+\tan B}{1-\tan A \tan B}=-\tan C$
$\Rightarrow \quad \frac{r^{2} / p q}{1-1}=-\tan C$
$\Rightarrow \quad \infty=-\tan C \Rightarrow C=90^{\circ}$
Hence, $\triangle A B C$ is a right angled triangle.
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