$\begin {array} {1 1} (A)\;\large\frac{a^2-b^2}{ab} & \quad (B)\;\large\frac{b^2-a^2}{2} \\ (C)\;\large\frac{b^2-a^2}{2ab} & \quad (D)\;\text{None of these} \end {array}$

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- Slope of a line $'m' = -\large\frac{coefficient \: of \: x}{coefficient \: of \: y}$
- Angle between two lines whose slopes are $m_1$ and $m_2$ is $ \tan \theta = \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$

Step 1:

Equation of the given lines are

$ \large\frac{x}{a}$$+\large\frac{y}{-b}$$=1$

$ \Rightarrow -bx+ay=-ab$----------(1) and

$ \large\frac{x}{b}$$+\large\frac{y}{-a}$$=1$

$ \Rightarrow -ax+by=-ab$--------(2)

Slope of the line (1) is $m_1=\large\frac{b}{a}$

Slope of the line (2) is $m_2=\large\frac{a}{b}$

$ \therefore \tan \theta = \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$

Now substituting the values we get,

$ \tan \theta = \bigg| \large\frac{\Large\frac{b}{a}-\Large\frac{a}{b}}{1+\Large\frac{b}{a} \times \Large\frac{a}{b}} \bigg|$

$ = \large\frac{b^2-a^2}{2ab}$

Hence c is the correct option.

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