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If $\sinh ^{-1}(\sqrt{8})+\sinh ^{-1}(\sqrt{24})=\alpha$, then $\sinh$ $\alpha=$
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Verified Answer
The correct answer is:
$6 \sqrt{6}+10 \sqrt{2}$
We have,
$\begin{array}{ll}
& \sin h^{-1}(\sqrt{8})+\sin h^{-1}(\sqrt{24})=\alpha \\
\text {Let } & \sin h^{-1}(\sqrt{8})=x \\
\Rightarrow & \sin h(x)=\sqrt{8} \\
& \cos (h x)=\sqrt{1+8}=\sqrt{9}=3 \\
\text {and } & \sin h^{-1}(\sqrt{24})=y \Rightarrow \sin h(y)=\sqrt{24} \\
& \cos h(y)=\sqrt{1+24}=5 \\
& \sin h(x+y)=\sin h x \cos h y
\end{array}$
$\begin{aligned}
\quad+\cos h x & \sin h y \\
\therefore \sin h(x+y) & =5 \sqrt{8}+3 \sqrt{24}=10 \sqrt{2}+6 \sqrt{6} \\
x+y & =\sin h^{-1}(6 \sqrt{6}+10 \sqrt{2}) \\
\alpha & =\sin h^{-1}(6 \sqrt{6}+10 \sqrt{2}) \\
\sin h(\alpha) & =6 \sqrt{6}+10 \sqrt{2}
\end{aligned}$
$\begin{array}{ll}
& \sin h^{-1}(\sqrt{8})+\sin h^{-1}(\sqrt{24})=\alpha \\
\text {Let } & \sin h^{-1}(\sqrt{8})=x \\
\Rightarrow & \sin h(x)=\sqrt{8} \\
& \cos (h x)=\sqrt{1+8}=\sqrt{9}=3 \\
\text {and } & \sin h^{-1}(\sqrt{24})=y \Rightarrow \sin h(y)=\sqrt{24} \\
& \cos h(y)=\sqrt{1+24}=5 \\
& \sin h(x+y)=\sin h x \cos h y
\end{array}$
$\begin{aligned}
\quad+\cos h x & \sin h y \\
\therefore \sin h(x+y) & =5 \sqrt{8}+3 \sqrt{24}=10 \sqrt{2}+6 \sqrt{6} \\
x+y & =\sin h^{-1}(6 \sqrt{6}+10 \sqrt{2}) \\
\alpha & =\sin h^{-1}(6 \sqrt{6}+10 \sqrt{2}) \\
\sin h(\alpha) & =6 \sqrt{6}+10 \sqrt{2}
\end{aligned}$
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