Question

prove that sinh^-1(tanx)=logtan(x/2+π /4)

Answer 1

Step-by-step explanation:

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Answer 2

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Question :-

If tan(x/2) = tanh(u/2), prove that

u = log tan(π/4 + u/2).

Proof :-

Given, tan(x/2) = tanh(u/2)

⇒ tan(x/2) =

$\frac{ sinh(u / 2 )}{cosh \: (u/2) }$

⇒tan(x/2) =

$\frac{e {}^{u} - e { }^{ - u/2}}{e {}^{u/2} + e {}^{ -u /2} }$

⇒ tan(x/2) = (eᵘ - 1)/(eᵘ + 1)

⇒ eᵘ - 1 = (eᵘ + 1) tan(x/2)

⇒ eᵘ - 1 = eᵘ tan(x/2) + tan(x/2)

⇒ {1 - tan(x/2)} eᵘ = 1 + tan(x/2)

⇒ eᵘ = {1 + tan(x/2)}/{1 - tan(x/2)}

⇒ eᵘ = tan(π/4 + x/2)

⇒ u = log tan(π/4 + x/2)

Hence, proved.

Rules :

• tanh(x) = {sinh(x)}/{cosh(x)}

• sinh(x) = (eˣ - e⁻ˣ)/2

• cosh(x) = (eˣ + e⁻ˣ)/2

• tan(A + B) = (tanA + tanB)/(1 - tanA tanB)

• eˣ = k ⇒ x = logₑk