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Question
Show that
`tan^-1(1/5) + tan^-1(1/7) + tan^-1(1/3) + tan^-1 (1/8) = pi/4.`
Solution
LHS = `tan^-1(1/5) + tan^-1(1/7) + tan^-1(1/3) + tan^-1 (1/8)`
`= tan^-1 [(1/5 + 1/7)/(1 - 1/5 xx 1/7)] + tan^-1 [(1/3+1/8)/(1 - 1/3 xx 1/8)]`
`= tan^-1 ((7 + 5)/(35 - 1)) + tan^-1 ((8 + 3)/(24 - 1))`
`= tan^-1 (12/34) + tan^-1 (11/23)`
`= tan^-1 (6/17) + tan^-1 (11/23)`
`= tan^-1 [(6/17 + 11/23)/(1 - 6/17 xx 11/23)]`
`= tan^-1((138 + 187)/(391 - 66)) = tan^-1 (325/325)`
`= tan^-1 (1) = tan^-1 (tan pi/4)`
`= pi/4`
= RHS.
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