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Question
Show that the tangent of an angle between the lines `x/a + y/b` = 1 and `x/a - y/b` = 1 is `(2ab)/(a^2 - b^2)`
Solution
Given that: `x/a + y/b` = 1 ......(i)
And `x/a - y/b` = 1 ......(ii)
Slope of equation (i) m1 (say) = `b/a`
And slope of equation (ii) m1 (say) = `b/a`
Let θ be the angle between the equation (i) and (ii)
∴ tan θ = `|(m_1 - m_2)/(1 + m_1m_2)|`
= `|(-b/a - b/a)/(1 + (- b/a)(b/a))|`
⇒ tan θ = `|(- (2b)/a)/(1 - b^2/a^2)|`
= `|(-2ab)/(a^2 - b^2)|`
⇒ tan θ = `(2ab)/(a^2 - b^2)`
Hence proved.
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