Question

Show that $\tanh \left(3x\right)=\frac{3\tanh x+\tan h^{3}x}{1+3\tan h^{2}x}$

Answer 1

$\tan h\left(3x\right)=\frac{3\tanh x+{\tanh }^{2}x}{1+3{\tanh }^{2}x}.$

$L.H.S=tanh\left(3x\right)$

$=-itan\left(3ix\right)$

$=-i\left[\frac{3\tan \left(ix\right)-{\tan }^2\left(ix\right)}{1-3{\tan }^2\left(ix\right)}\right]$

$=-i\left[\frac{3i\tanh x-i^3{\tanh }^{3}x}{1-3i^2{\tanh }^{2}x}\right]$

$=\frac{3\tanh x+{\tanh }^{2}x}{1+3{\tanh }^{2}x}$

Hence, we get.

L.H.S=R.H.S