operatorname{sech}^2left(tanh ^{-1} frac{1}{2 | vedclass

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(D) We know the identities for inverse hyperbolic functions: $	anh^{-1}(x) = \operatorname{sech}^{-1}\left(\sqrt{1-x^2}ight)$ and $\coth^{-1}(x) =

Vedclass · Accepted Answer

$\frac{35}{4}$

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(D) We know the identities for inverse hyperbolic functions: $	anh^{-1}(x) = \operatorname{sech}^{-1}\left(\sqrt{1-x^2}ight)$ and $\coth^{-1}(x) = \operatorname{cosech}^{-1}\left(\frac{1}{\sqrt{x^2-1}}ight)$. For the first term: $	anh^{-1}\left(\frac{1}{2}ight) = \operatorname{sech}^{-1}\left(\sqrt{1-(\frac{1}{2})^2}ight) = \operatorname{sech}^{-1}\left(\sqrt{\frac{3}{4}}ight) = \operatorname{sech}^{-1}\left(\frac{\sqrt{3}}{2}ight)$. Thus,$\operatorname{sech}^2\left(	anh^{-1} \frac{1}{2}ight) = \left(\frac{\sqrt{3}}{2}ight)^2 = \frac{3}{4}$. For the second term: $\coth^{-1}(3) = \operatorname{cosech}^{-1}\left(\frac{1}{\sqrt{3^2-1}}ight) = \operatorname{cosech}^{-1}\left(\frac{1}{\sqrt{8}}ight)$. Thus,$\operatorname{cosech}^2\left(\coth^{-1} 3ight) = \left(\frac{1}{\sqrt{8}}ight)^{-2} = 8$. Adding both terms: $\frac{3}{4} + 8 = \frac{3+32}{4} = \frac{35}{4}$.

$\operatorname{sech}^2\left(\tanh ^{-1} \frac{1}{2}\right)+\operatorname{cosech}^2\left(\operatorname{coth}^{-1} 3\right)=$

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