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

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(D) 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

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

Vedclass · Answer

(D) 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)$ $\coth^{-1}(-2) = \frac{1}{2} \ln\left(\frac{-2+1}{-2-1}ight) = \frac{1}{2} \ln\left(\frac{-1}{-3}ight) = \frac{1}{2} \ln\left(\frac{1}{3}ight) = -\frac{1}{2} \ln(3)$ Now, sum them up: $\ln(2+\sqrt{5}) + \ln(2+\sqrt{3}) - \frac{1}{2} \ln(5) - \frac{1}{2} \ln(3)$ $= \ln((2+\sqrt{5})(2+\sqrt{3})) - \frac{1}{2} \ln(15)$ $= \ln((2+\sqrt{5})(2+\sqrt{3})) - \ln(\sqrt{15})$ $= \ln\left(\frac{(2+\sqrt{5})(2+\sqrt{3})}{\sqrt{15}}ight)$ $= \ln\left(\frac{(2+\sqrt{5})(2+\sqrt{3}) \sqrt{3}}{\sqrt{5} \cdot 3} \cdot \sqrt{3}ight) = \ln\left(\frac{(2+\sqrt{5})(2+\sqrt{3}) \sqrt{3}}{\sqrt{5}}ight)$ Thus, the correct option is $D$.

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

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