If e^{left(sinh ^{-1} 2+cosh ^{-1} sqrt{6}rig | vedclass

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Question

(A) Let $x = \sinh^{-1} 2$ and $y = \cosh^{-1} \sqrt{6}$.We know that $\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1})$, so $x = \ln(2 + \sqrt{2^2 + 1}) = \ln(

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(A) Let $x = \sinh^{-1} 2$ and $y = \cosh^{-1} \sqrt{6}$.We know that $\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1})$, so $x = \ln(2 + \sqrt{2^2 + 1}) = \ln(2 + \sqrt{5})$.We know that $\cosh^{-1} x = \ln(x + \sqrt{x^2 - 1})$, so $y = \ln(\sqrt{6} + \sqrt{6 - 1}) = \ln(\sqrt{6} + \sqrt{5})$.Then $e^{x+y} = e^x \cdot e^y = (2 + \sqrt{5})(\sqrt{6} + \sqrt{5})$.Expanding this, we get $2\sqrt{6} + 2\sqrt{5} + \sqrt{30} + 5 = 5 + 2\sqrt{6} + 2\sqrt{5} + \sqrt{30}$.Comparing this with the form $(a+(b+\sqrt{c}) \sqrt{a}+b \sqrt{c})$, we observe that the expression is $5 + 2\sqrt{6} + 2\sqrt{5} + \sqrt{30}$.Wait, let us re-evaluate the expression: $(2 + \sqrt{5})(\sqrt{6} + \sqrt{5}) = 2\sqrt{6} + 2\sqrt{5} + \sqrt{30} + 5$.Given the form $(a+(b+\sqrt{c}) \sqrt{a}+b \sqrt{c})$, let $a=5, b=2, c=6$. Then $5 + (2+\sqrt{6})\sqrt{5} + 2\sqrt{6} = 5 + 2\sqrt{5} + \sqrt{30} + 2\sqrt{6}$.This matches our result. Thus $a=5, b=2, c=6$.Therefore, $a+b+c = 5+2+6 = 13$.

If $e^{\left(\sinh ^{-1} 2+\cosh ^{-1} \sqrt{6}\right)}=(a+(b+\sqrt{c}) \sqrt{a}+b \sqrt{c})$, then $a+b+c=$

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