sinh^{-1}(2) + cosh^{-1}(2) - tanh^{-1}left(f | vedclass

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Question

(A) We use the logarithmic definitions of inverse hyperbolic functions:$\sinh^{-1}(x) = \ln(x + \sqrt{x^2+1})$$\cosh^{-1}(x) = \ln(x + \sqrt{x^2-1})$$

Vedclass · Accepted Answer

$\ln \left(\frac{(2+\sqrt{3})(2+\sqrt{5})}{\sqrt{15}}\right)$

Vedclass · Answer

(A) We use the logarithmic definitions of inverse hyperbolic functions:$\sinh^{-1}(x) = \ln(x + \sqrt{x^2+1})$$\cosh^{-1}(x) = \ln(x + \sqrt{x^2-1})$$	anh^{-1}(x) = \frac{1}{2} \ln\left(\frac{1+x}{1-x}ight)$$\coth^{-1}(x) = \frac{1}{2} \ln\left(\frac{x+1}{x-1}ight)$Substituting the values:$\sinh^{-1}(2) = \ln(2 + \sqrt{5})$$\cosh^{-1}(2) = \ln(2 + \sqrt{3})$$	anh^{-1}\left(\frac{2}{3}ight) = \frac{1}{2} \ln\left(\frac{1+2/3}{1-2/3}ight) = \frac{1}{2} \ln(5)$$\coth^{-1}(-2) = \frac{1}{2} \ln\left(\frac{-2+1}{-2-1}ight) = \frac{1}{2} \ln\left(\frac{-1}{-3}ight) = \frac{1}{2} \ln\left(\frac{1}{3}ight) = -\frac{1}{2} \ln(3)$Summing them up:$\ln(2+\sqrt{5}) + \ln(2+\sqrt{3}) - \frac{1}{2} \ln(5) - \frac{1}{2} \ln(3)$$= \ln((2+\sqrt{5})(2+\sqrt{3})) - \frac{1}{2} \ln(15)$$= \ln((2+\sqrt{5})(2+\sqrt{3})) - \ln(\sqrt{15})$$= \ln\left(\frac{(2+\sqrt{5})(2+\sqrt{3})}{\sqrt{15}}ight)$Thus, the correct option is $A$.

$\sinh^{-1}(2) + \cosh^{-1}(2) - \tanh^{-1}\left(\frac{2}{3}\right) + \coth^{-1}(-2) = $

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