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