Answer to Question #347753 in Algebra for Xhaka

Question #347753

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

1
Expert's answer
2022-06-06T05:32:32-0400

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


cos2x+sin2x=1=>ch2xsh2x=1\cos^2x+\sin^2x=1=>\ch^2x-\sh^2x=1

Then


1tanh2x=1sh2xch2x=ch2xsh2xch2x1-\tanh^2x=1-\dfrac{\sh ^2x}{\ch^2x}=\dfrac{\ch^2 x-\sh^2 x}{\ch^2 x}

=1ch2x=sech2x=\dfrac{1}{\ch^2x}=\text{sech}^2x


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