e^{left(sec h^{-1} frac{1}{2}+tan h^{-1} frac | vedclass

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Question

(B) We know the logarithmic forms of inverse hyperbolic functions:$\sec h^{-1} x = \log \left(\frac{1+\sqrt{1-x^2}}{x}ight)$$	an h^{-1} x = \frac{1

Vedclass · Accepted Answer

$\frac{3+2 \sqrt{3}+3 \sqrt{5}+2 \sqrt{15}}{2}$

Vedclass · Answer

(B) We know the logarithmic forms of inverse hyperbolic functions:$\sec h^{-1} x = \log \left(\frac{1+\sqrt{1-x^2}}{x}ight)$$	an h^{-1} x = \frac{1}{2} \log \left(\frac{1+x}{1-x}ight)$$\sin h^{-1} x = \log \left(x+\sqrt{x^2+1}ight)$For $x = \frac{1}{2}$:$\sec h^{-1} \frac{1}{2} = \log \left(\frac{1+\sqrt{3/4}}{1/2}ight) = \log (2+\sqrt{3})$$	an h^{-1} \frac{1}{2} = \frac{1}{2} \log \left(\frac{3/2}{1/2}ight) = \frac{1}{2} \log 3 = \log \sqrt{3}$$\sin h^{-1} \frac{1}{2} = \log \left(\frac{1}{2} + \sqrt{\frac{1}{4}+1}ight) = \log \left(\frac{1+\sqrt{5}}{2}ight)$Now,the expression becomes:$e^{\log (2+\sqrt{3}) + \log \sqrt{3} + \log \left(\frac{1+\sqrt{5}}{2}ight)}$$= e^{\log \left((2+\sqrt{3}) \cdot \sqrt{3} \cdot \frac{1+\sqrt{5}}{2}ight)}$$= (2\sqrt{3}+3) \cdot \left(\frac{1+\sqrt{5}}{2}ight)$$= \frac{2\sqrt{3} + 2\sqrt{15} + 3 + 3\sqrt{5}}{2}$$= \frac{3 + 2\sqrt{3} + 3\sqrt{5} + 2\sqrt{15}}{2}$

$e^{\left(\sec h^{-1} \frac{1}{2}+\tan h^{-1} \frac{1}{2}+\sin h^{-1} \frac{1}{2}\right)}=$

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