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Q. The angle between asymptotes of $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$

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Solution:

$\theta=2 \tan ^{-1}\left(\frac{b}{a}\right)=2 \tan ^{-1}\left(\frac{3}{4}\right)$
$=\tan ^{-1}\left(\frac{2(3 / 4)}{1-9 / 16}\right)=\tan ^{-1}\left(\frac{3 / 2}{7 / 16}\right)$
$\therefore \theta=\tan ^{-1}\left(\frac{24}{7}\right)$

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