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Question
Prove that `tan^-1 (m/n) - tan^-1 ((m - n)/(m + n)) = pi/4`
Sum
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Solution
LHS = `tan^-1 (m/n) - tan^-1 ((m - n)/(m + n))`
`= tan^-1 ((m/n - (m-n)/(m+n))/(1 + (m/n)((m-n)/(m+n))))`
`= tan^-1 (((m(m + n) - n(m - n))/(n(m+n)))/((n(m+n)+m(m-n))/(n(m+n))))`
`= tan^-1 ((m^2 + mn - nm + n^2)/(nm + n^2 + m^2 - mn))`
`= tan^-1 ((m^2 + n^2)/(m^2 + n^2))`
`= tan^-1 (1) = pi/4`
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