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If $\tan \mathrm{A}=\mathrm{x}+1$ and $\tan \mathrm{B}=\mathrm{x}-1$, then $\mathrm{x}^{2} \tan (\mathrm{A}-\mathrm{B})$ has the
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The correct answer is:
2
$x^{2} \tan (A-B)$
$=x^{2}\left(\frac{\tan A-\tan B}{1+\tan A \tan B}\right)=x^{2}\left(\frac{(x+1)-(x-1)}{1+\left(x^{2}-1\right)}\right)$
$=x^{2}\left(\frac{2}{x^{2}}\right)=2$
$=x^{2}\left(\frac{\tan A-\tan B}{1+\tan A \tan B}\right)=x^{2}\left(\frac{(x+1)-(x-1)}{1+\left(x^{2}-1\right)}\right)$
$=x^{2}\left(\frac{2}{x^{2}}\right)=2$
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