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Question
Show that `cot^-1 1/3 - tan^-1 1/3 = cot^-1 3/4`.
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Solution
LHS = `cot^-1 1/3 - tan^-1 1/3`
`= tan^-1 3 - tan^-1 1/3 ....[because cot^-1 "x" = tan^-1 (1/"x")]`
`= tan^-1 [(3 - 1/3)/(1 + 3(1/3))]`
`= tan^-1 [(8/3)/(1 + 1)]`
`= tan^-1 (4/3)`
`= cot^-1 (3/4) ....[tan^-1 "x" = cot^-1 (1/"x")]`
= RHS.
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