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Question
If tan x = `"n"/("n" + 1)` and tan y = `1/(2"n" + 1)`, find tan(x + y)
Solution
tan x = `"n"/("n" + 1)`, tan y = `1/(2"n" + 1)`
tan(x + y) = `(tanx + tany)/(1 - tanx tany)`
= `("n"/("n" + 1) + 1/(2"n" + 1))/(1 - "n"/("n" + 1) * 1/(2"n" + 1))`
= `(("n"(2"n" + 1) + "n" + 1)/(("n" + 1)(2"n" + 1)))/((("n" + 1)(2"n" + 1) - "n")/(("n" + 1)(2"n" + 1))`
= `("n"(2"n" + 1) + "n" + 1)/(("n" + 1)(2"n" + 1) - "n")`
= `(2"n"^2 + "n" + "n" + 1)/(2"n"^2 + "n" + 2"n" + 1 - "n")`
= `(2"n"^2 + 2"n" + 1)/(2"n"^2 + 2"n" + 1)`
tan(x + y) = 1
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