If theta is the acute angle between the lines | vedclass

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(D) Given lines are $L_1: \frac{x}{a} + \frac{y}{b} = 1$ and $L_2: \frac{x}{b} + \frac{y}{a} = 1$.Slope of $L_1$ is $m_1 = -\frac{b}{a}$.Slope of $L_2

Vedclass · Accepted Answer

$\left|\frac{a^2-b^2}{a^2+b^2}\right|$

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(D) Given lines are $L_1: \frac{x}{a} + \frac{y}{b} = 1$ and $L_2: \frac{x}{b} + \frac{y}{a} = 1$.Slope of $L_1$ is $m_1 = -\frac{b}{a}$.Slope of $L_2$ is $m_2 = -\frac{a}{b}$.The angle $	heta$ between two lines is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.Substituting the slopes: $	an 	heta = \left| \frac{-\frac{b}{a} - (-\frac{a}{b})}{1 + (-\frac{b}{a})(-\frac{a}{b})} ight| = \left| \frac{\frac{a}{b} - \frac{b}{a}}{1 + 1} ight| = \left| \frac{a^2 - b^2}{2ab} ight|$.Since $	an 	heta = \frac{	ext{opposite}}{	ext{adjacent}} = \frac{|a^2 - b^2|}{|2ab|}$,the hypotenuse is $\sqrt{(a^2 - b^2)^2 + (2ab)^2} = \sqrt{a^4 + b^4 - 2a^2b^2 + 4a^2b^2} = \sqrt{(a^2 + b^2)^2} = |a^2 + b^2|$.Therefore,$\sin 	heta = \frac{	ext{opposite}}{	ext{hypotenuse}} = \left| \frac{a^2 - b^2}{a^2 + b^2} ight|$.

If $\theta$ is the acute angle between the lines $\frac{x}{a}+\frac{y}{b}=1$ and $\frac{x}{b}+\frac{y}{a}=1$,then $\sin \theta=$

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