Q.Starting from the definitions of $\tanh x$ and $\operatorname{sech} x$ in terms of $\sinh x$ and $\cosh x$, prove that $1 - \tanh^2 x = \operatorname{sech}^2 x$.

Factor the difference of squares $\cosh^4x-\sinh^4x=(\cosh^2x-\sinh^2x)(\cosh^2x+\sinh^2x)$; the first factor is $1$ (hyperbolic Pythagorean identity) and the second factor is exactly $\cosh(2x)$.

$\cosh^4x - \sinh^4x = (\cosh^2x-\sinh^2x)(\cosh^2x+\sinh^2x)$

Using the hyperbolic identity $\cosh^2x-\sinh^2x = 1$:

$= 1\cdot(\cosh^2x+\sinh^2x) = \cosh^2x+\sinh^2x$

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Use the standard log form $\tanh^{-1}x=\frac12\ln\frac{1+x}{1-x}$ and substitute $x=\frac12$.

Step 1 — State the formula.

$\tanh^{-1}x = \frac12\log_e\left(\frac{1+x}{1-x}\right), \quad -1<x<1$

Step 2 — Substitute $x=\dfrac12$. …

$\cosh x - \sinh x$ simplifies exactly to $e^{-x}$ by definition, which makes the whole identity a one-line exponent rule.

By definition, $\cosh x = \dfrac{e^x+e^{-x}}{2}$ and $\sinh x = \dfrac{e^x-e^{-x}}{2}$.

$\cosh x - \sinh x = \dfrac{e^x+e^{-x}}{2} - \dfrac{e^x-e^{-x}}{2} = \dfrac{2e^{-x}}{2} = e^{-x}$

So $(\cosh x-\sinh x)^n = (e^{-x})^n = e^{-nx}$

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