Question

Prove whether the following are identities 2tanh 1/2x / 1−tanh^2 1/2 x = sinh x​

Answer 1

Recall that

$\cosh^2(x) - \sinh^2(x) = 1$

Dividing both sides by cosh²(x) gives

$1 - \tanh^2(x) = \mathrm{sech}^2(x)$

Also, recall the identity

$\sinh(2x) = 2\sinh(x)\cosh(x)$

Then

$\dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = \dfrac{2\tanh\left(\frac x2\right)}{\mathrm{sech}^2\left(\frac x2\right)} \\\\ \dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = 2\tanh\left(\dfrac x2\right)\cosh^2\left(\dfrac x2\right) \\\\ \dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = 2\sinh\left(\dfrac x2\right)\cosh\left(\dfrac x2\right) \\\\\dfrac{2\tanh\left(\frac x2\right)}{1 - \tanh^2\left(\frac x2\right)} = \sinh(x)$