Question

If ${\sinh }^{-1}(\sqrt{8})+{\sinh }^{-1}(\sqrt{24})=\alpha$, then sinh $\alpha =$

• A. $6\sqrt{6}-10\sqrt{2}$
• B. $6\sqrt{6}+10\sqrt{2}$
• C. $16\sqrt{6}$
• D. $16\sqrt{6}+4\sqrt{2}$

Answer 1

Answer: (B)

We have,
$\sin h^{-1}(\sqrt{8})+\sin h^{-1}(\sqrt{24})=\alpha$
Let $\sin h^{-1}(\sqrt{8})=x$
$\Rightarrow \sin h(x)=\sqrt{8}$
$\cos (hx)=\sqrt{1+8}=\sqrt{9}=3$
and $\sin h^{-1}(\sqrt{24})=y\Rightarrow \sin h(y)=\sqrt{24}$
$\therefore \cos h(y)=\sqrt{1+24}=5$
$\sin h(x+y)=\sin hx\cos hy$
$\sin h(x+y)=5\sqrt{8}+3\sqrt{24}=10\sqrt{2}+6\sqrt{6}$
$x+y=\cos hx\sin hy$
$\sin h^{-1}(6\sqrt{6}+10\sqrt{2})$
$\alpha =\sin h^{-1}(6\sqrt{6}+10\sqrt{2})$
$\sin h(\alpha )=6\sqrt{6}+10\sqrt{2}$