A) \[{{\tan }^{-1}}\frac{{{a}^{2}}-{{b}^{2}}}{ab}\]
B) \[{{\tan }^{-1}}\frac{{{b}^{2}}-{{a}^{2}}}{2}\]
C) \[{{\tan }^{-1}}\frac{{{b}^{2}}-{{a}^{2}}}{2ab}\]
D) None of these
Correct Answer: C
Solution :
Equation of lines are \[\frac{x}{a}-\frac{y}{b}=1\]and \[\frac{x}{b}-\frac{y}{a}=1\] Þ \[{{m}_{1}}=\frac{b}{a}\]and \[{{m}_{2}}=\frac{a}{b}\] Therefore \[\theta ={{\tan }^{-1}}\frac{\frac{b}{a}-\frac{a}{b}}{1+\frac{b}{a}.\frac{a}{b}}={{\tan }^{-1}}\frac{{{b}^{2}}-{{a}^{2}}}{2ab}\].
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