Q.355. Trigonometry proofs

Question 355.
If tan x/2 = tanh y/2, then  prove that
(a) sinh y = tan x and
(b) y=log tan(π/4+x/2).

Answer 355.
(a)
tanx
= 2tan(x/2) / [1 - tan^2 (x/2)]
= 2 tanh (y/2) / [1 - tanh^2 (y/2)]
= 2 [{(e^(y/2) - e^(-y/2)} / {e^(y/2) + e^(-y/2)}] / [1 - [{e^(y2) - e^(-y/2)} / {e^(y/2) + e^(y/2)}]^2 ]
= 2 [{e^(y/2) - e^(-y/2)} * {e^(y/2) + e^(-y/2)}] / [{e^(y/2) + e^(-y/2)}^2 - {e^(y/2) - e^(-y/2)}^2]
= 2 (e^y - e^-y) / 4
= ((e^y - e^-y) / 2
= sinh y

(b)
tan(x/2) = tanh (y/2)

tan (π/4+x/2).
= [1 + tan(x/2)] / [1 - tan(x/2)]
= [1 + tanh (y/2)] / [1 - tanh (y/2)]
= [1 + {e^(y/2) - e^(-y/2)} / {e^(y/2) + e^(-y/2)}] / [1 - {e^(y/2) - e^(-y/2)} / {e^(y/2) + e^(-y/2)}]
= [2e^(y/2) / {e^(y/2) + e^(-y/2)}] / [2e^(-y/2) / {e^(y/2) + e^(-y/2)}]
= e^y
=> tan(π/4+x/2) = e^y
=> y = log tan (π/4+x/2).

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