Question

Prove that $\frac{tanh\ x}{sech\ x\ -\ 1} + \frac{tanh\ x}{sech\ x\ +\ 1} = 2 cosech\ x, \ for \ x\ \neq\ 0$.

Answer 1

$given \\ \\ \frac{tanh(x)}{sech(x) - 1} + \frac{tanh(x)}{sech(x) + 1} \\ = \frac{tanhx}{sechx - 1} \times \frac{sechx + 1}{sechx + 1} + \frac{tanhx}{sechx + 1} \times \frac{sechx - 1}{sechx - 1} \\ = \frac{tanhx}{ {sech}^{2}x - 1 } (sechx + 1) + \frac{tanhx}{ {sech}^{2}x - 1 } (sechx - 1) \\ < since \: {sech}^{2} x - 1 = {tanh}^{2} x > \\ = \frac{sechx + 1}{tanhx} + \frac{sechx - 1}{tanhx} \\ = \frac{2sechx}{tanhx} \\ = \frac{2sechx}{sinhx} coshx \\ = 2cosechx$
hope it helps