Question

$sin{h}^{-1}2+sin{h}^{-1}3=x$, then $coshx$ will be

• A. $\frac{3\sqrt{5}+2\sqrt{1}0}{2}$
• B. $\frac{3\sqrt{5}-2\sqrt{1}0}{2}$
• C. $\frac{12+2\sqrt{5}0}{2}$
• D. $\frac{12-2\sqrt{5}0}{2}$

Answer 1

Answer: (C)

$\frac{12+2\sqrt{5}0}{2}$

${\sinh }^{-1}2+{\sinh }^{-1}3=x$

$\because {\sinh }^{-1}2=\log (x+\sqrt{1+x^2})$

${\sinh }^{-1}2=\log (2+\sqrt{1+4})$

$=\log (2+\sqrt{5})$

${\sinh }^{-1}3=\log (3+\sqrt{1+3^2})$

$=\log (3+\sqrt{10})$

${\sinh }^{-1}2+{\sinh }^{-1}3=\log (2+\sqrt{5})+\log (3+\sqrt{10})$

$=\log \left(\right(2+\sqrt{5})\times (3+\sqrt{10}\left)\right)=x$

$\Rightarrow 6+2\sqrt{10}+3\sqrt{5}+\sqrt{5}\sqrt{10}=e^x$

$\because \cosh x=\frac{e^x+e^{-x}}{2}$

$=\frac{e^x+\frac{1}{e^x}}{2}$

$=\frac{(2+\sqrt{5})+(3+\sqrt{10})+\frac{(2+\sqrt{5})+(3+\sqrt{10})}{(2+\sqrt{5})+(3+\sqrt{10})\cdot (2+\sqrt{5})+(3+\sqrt{10})}}{2}$

$=\frac{(2+\sqrt{5})+(3+\sqrt{10})+6+2\sqrt{10}-3\sqrt{5}+\sqrt{50}}{2}$

$=\frac{(6+\sqrt{50})2}{2}=(6+\sqrt{50})$