Question

If tan A – tan B = x and cot B – cot A = y prove that cot(A – B) = `1/x + 1/y`.

Answer 1

`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.