Q.121. Trigonometry.

Question 121.
If tanh(x/2) = tan(x/2),proove that cosx coshx = 1.

Answer 121.
tanh(x/2) =  tan(x/2)
=> tan^2 h(x/2) = tan^2 (x/2)
=> (cos hx - 1) / (cos hx + 1) = (1 - cosx) / (1 + cosx)
=> (1 + cosx)(cos hx - 1) = (1 - cosx)(cos hx + 1)
=> cos hx + cosx cos hx - 1 - cosx
= cos hx - cosx cos hx + 1 - cosx
=> 2 cosx cos hx = 2
=> cosx cos hx = 1.

Alternate method:
cosh (x/2) = (1/2) a, where a = e^(x/2) + e^(-x/2)
sinh (x/2) = (1/2) b, where b = e^(x/2) - e^(-x/2)

=> a^2 = e^(x) + 2 + e^(-x) and b^2 = e^(x) - 2 + e^(-x)
=> (a^2 - b^2) / (a^2 + b^2) = 2 / [(e^(x) + e^(-x)]
=> (a^2 - b^2) / (a^2 + b^2) = 1 / cosh x ... ( 1 )

Now, tanh (x/2) = tan(x/2)
=> tanh^2 (x/2) = tan^ (x/2) = (1 - cosx) / (1 + cosx)
=> sinh^2 (x/2) / cosh^ (x/2) = (1 - cosx) / (1 + cosx)
=> b^2 / a^2 = (1 - cosx) / (1 + cosx)

By componendo et dividendo,
(a^2 - b^2) / (a^2 + b^2) = cosx
=> 1 / cosh x = cosx (using result ( 1 ) above)
=> cosx cosh x = 1.

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