Prove that 𝒕𝒂𝒏𝒉−𝟏(𝒔𝒊𝒏 𝜽) = 𝒄𝒐𝒔𝒉−𝟏(𝒔𝒆𝒄 𝜽)

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#### Solution

`L.H.S = tanh^1(sin θ)`

We know that, `tanh^-1(x)=1/2log ((1+x)/(1-x))`

∴ `L.H.S=1/2 log ((1+sin θ)/(1-sin θ) )`

`R.H.S=cosh^-1 (secθ)`

We know that , `cosh^-1 (x)=log (x+sqrt(x^2-1))`

∴ R.H.S =` log(sec θ +sqrt(sec^2θ-1))`

= `log (1/cos θ+sin θ/cos θ)` ...........`{sqrt(sec^2θ-1)=tanθ=sinθ/cosθ}`

=` log ((1+sin θ)/cos θ)`

=` log (1+sin θ/sqrt(1-sin^2θ))`

=` log (sqrt(1+sinθ)/sqrt(1-sin θ))`

= `1/2 log ((1+ sin θ)/(1-sin θ))`

∴ `tanh^-1(sin θ)=cosh^-1(sec θ)`

Hence Proved.

Concept: .Circular Functions of Complex Number

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