Question

Proof using Osborne's rule that 1-tanh^2x=Sech^2x

Answer 1

The rule states that one replaces every occurrence of sine or cosine with the corresponding hyperbolic sine or cosine, and wherever one has a product of two sines, the product of the hyperbolic sines must be negated.

By the Osborne's rule

$$\cos^2x+\sin^2x=1=>\ch^2x-\sh^2x=1$$

Then

$$1-\tanh^2x=1-\dfrac{\sh ^2x}{\ch^2x}=\dfrac{\ch^2 x-\sh^2 x}{\ch^2 x}$$

$$=\dfrac{1}{\ch^2x}=\text{sech}^2x$$