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Angle between the lines $\frac{x}{a}+\frac{y}{b}=1$ and $\frac{x}{a}-\frac{y}{b}=1$ is
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Verified Answer
The correct answer is:
$2 \tan ^{-1} \frac{b}{a}$
The lines are $b x+a y-a b=0$ and $b x-a y-a b=0$
Hence the required angle is
$\tan ^{-1}\left|\frac{a b-(-a b)}{b^2+\left(-a^2\right)}\right|=\tan ^{-1}\left|\frac{2 a b}{b^2-a^2}\right|=2 \tan ^{-1} \frac{b}{a}$
Hence the required angle is
$\tan ^{-1}\left|\frac{a b-(-a b)}{b^2+\left(-a^2\right)}\right|=\tan ^{-1}\left|\frac{2 a b}{b^2-a^2}\right|=2 \tan ^{-1} \frac{b}{a}$
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