coth^{-1}(2) + operatorname{cosech}^{-1}(-2sq | vedclass

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Question

(A) Given expression: $\coth^{-1}(2) + \operatorname{cosech}^{-1}(-2\sqrt{2})$.Since $\operatorname{cosech}^{-1}(-x) = -\operatorname{cosech}^{-1}(x)$

Vedclass · Accepted Answer

$\log \sqrt{\frac{3}{2}}$

Vedclass · Answer

(A) Given expression: $\coth^{-1}(2) + \operatorname{cosech}^{-1}(-2\sqrt{2})$.Since $\operatorname{cosech}^{-1}(-x) = -\operatorname{cosech}^{-1}(x)$, the expression becomes $\coth^{-1}(2) - \operatorname{cosech}^{-1}(2\sqrt{2})$.Using the formula $\coth^{-1}(x) = \frac{1}{2} \log \left(\frac{x+1}{x-1}\right)$ for $|x| > 1$:$\coth^{-1}(2) = \frac{1}{2} \log \left(\frac{2+1}{2-1}\right) = \frac{1}{2} \log(3) = \log \sqrt{3}$.Using the formula $\operatorname{cosech}^{-1}(x) = \log \left(\frac{1 + \sqrt{x^2+1}}{x}\right)$ for $x > 0$:$\operatorname{cosech}^{-1}(2\sqrt{2}) = \log \left(\frac{1 + \sqrt{(2\sqrt{2})^2 + 1}}{2\sqrt{2}}\right) = \log \left(\frac{1 + \sqrt{8+1}}{2\sqrt{2}}\right) = \log \left(\frac{1+3}{2\sqrt{2}}\right) = \log \left(\frac{4}{2\sqrt{2}}\right) = \log \sqrt{2}$.Substituting these values back into the expression:$\log \sqrt{3} - \log \sqrt{2} = \log \left(\frac{\sqrt{3}}{\sqrt{2}}\right) = \log \sqrt{\frac{3}{2}}$.

$\coth^{-1}(2) + \operatorname{cosech}^{-1}(-2\sqrt{2}) = $

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