Question

Prove that `cos^-1tanh(log x)+ = π – 2(x-x^3/3+x^5/5.........)`

Answer 1

We know, tanhθ=` sinθ/coshθ`

=`((e^θ-e^-θ)/2)/((e^θ+e^-θ)/2)`

∴ `tanhθ=(e^θ-e^-θ)/(e^θ+e^-θ)`

Put θ=log x, 

∴ `tanh(logx)=(e^log-e^logx)/(e^log+e^-logx)`

=`(e^logx-e^(logx^-1))/(e^log+e^(logx^-1))`

=`(x-x^-1)/(x+x^-1)`

=`(x(1-x^-2))/(x(1+x^-2))` 

Let` y = cos^-1[tanh(log x)]` 

= `cos^-1[(1-x^-2)/(1+x^-2)]`

= `cos^-1[(1-(x^1)^-2)/(1+(x^-1)^2)]`

Put `x^-1=tanθ`

∴ `y = cos^-1 ((1-tan^2θ)/(1+tan^2θ))`

= `cos^-1(cos2θ)`
=` 2θ` 

=`2 tan^-1(1/x)`                          (From 1) 

= `2cot^-1x`

= `2(pi/2-tan^-1x)`

=`pi-2tan^-1x`

=`pi-2(x-x^3/3+x^5/5........)`

Hence, `cos^-1tanh(log x) = π – 2(x-x^3/3+x^5/5..........)`