operatorname{coth}^2 x - tanh^2 x = | vedclass

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Question

(B) We know that $\operatorname{coth} x = \frac{\cosh x}{\sinh x}$ and $	anh x = \frac{\sinh x}{\cosh x}$.Given expression: $\operatorname{coth}^2 x

Vedclass · Accepted Answer

$4 \operatorname{sech} 2x \operatorname{coth} 2x$

Vedclass · Answer

(B) We know that $\operatorname{coth} x = \frac{\cosh x}{\sinh x}$ and $	anh x = \frac{\sinh x}{\cosh x}$.Given expression: $\operatorname{coth}^2 x - 	anh^2 x = \frac{\cosh^2 x}{\sinh^2 x} - \frac{\sinh^2 x}{\cosh^2 x}$.Taking $LCM$: $\frac{\cosh^4 x - \sinh^4 x}{\sinh^2 x \cosh^2 x}$.Using $a^2 - b^2 = (a-b)(a+b)$,we get $\frac{(\cosh^2 x - \sinh^2 x)(\cosh^2 x + \sinh^2 x)}{\sinh^2 x \cosh^2 x}$.Since $\cosh^2 x - \sinh^2 x = 1$ and $\cosh^2 x + \sinh^2 x = \cosh 2x$,the expression becomes $\frac{\cosh 2x}{\sinh^2 x \cosh^2 x}$.Multiply numerator and denominator by $4$: $\frac{4 \cosh 2x}{4 \sinh^2 x \cosh^2 x} = \frac{4 \cosh 2x}{(2 \sinh x \cosh x)^2} = \frac{4 \cosh 2x}{\sinh^2 2x}$.This can be written as $4 \cdot \frac{\cosh 2x}{\sinh 2x} \cdot \frac{1}{\sinh 2x} = 4 \operatorname{coth} 2x \operatorname{cosech} 2x$.

$\operatorname{coth}^2 x - \tanh^2 x =$

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