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Question
If tan A – tan B = x and cot B – cot A = y prove that cot(A – B) = `1/x + 1/y`.
Solution
`1/x + 1/y = 1/(tan "A" - tan "B") + 1/(cot "B" - cot "A")`
= `1/(tan "A" - tan "B") + 1/(1/tan "B" - 1/(tan "A"))`
= `1/(tan "A" - tan "B") + 1/(((tan "A" - tan "B")/(tan "A" tan "B")))`
= `1/(tan "A" - tan "B") + (tan "A" tan "B")/(tan "A" - tan "B") = (1 + tan "A" tan "B")/(tan "A" - tan "B")`
`= 1/(tan ("A - B")) ....(because tan ("A - B") = (tan "A" - tan "B")/(1 - tan "A" tan "B"))`
= cot (A - B) = LHS
Hence proved.
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