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If $\operatorname{sech}^{-1}\left(\frac{1}{2}\right)-\operatorname{cosec~h}^{-1}\left(\frac{3}{4}\right)=\log _e k$, then
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Verified Answer
The correct answer is:
$9 k^2-12 k+1=0$
Given, $\operatorname{sech}^{-1}\left(\frac{1}{2}\right)-\operatorname{cosech}^{-1}\left(\frac{3}{4}\right)=\log _e k$
$\Rightarrow \log _e\left(\sqrt{2^2-1}+2\right)-\log _e\left(\sqrt{\left(\frac{4}{3}\right)^2}+1+\frac{4}{3}\right)=\log _e k$
$\Rightarrow \log _e(\sqrt{3}+2)-\log \left(\frac{5}{3}+\frac{4}{3}\right)=\log _e k$
$\Rightarrow \quad \log _e\left(\frac{\sqrt{3}+2}{3}\right)=\log _e k \Rightarrow 3 k-2=\sqrt{3}$
$\Rightarrow 9 k^2-12 k+4=3 \Rightarrow 9 k^2-12 k+1=0$
$\Rightarrow \log _e\left(\sqrt{2^2-1}+2\right)-\log _e\left(\sqrt{\left(\frac{4}{3}\right)^2}+1+\frac{4}{3}\right)=\log _e k$
$\Rightarrow \log _e(\sqrt{3}+2)-\log \left(\frac{5}{3}+\frac{4}{3}\right)=\log _e k$
$\Rightarrow \quad \log _e\left(\frac{\sqrt{3}+2}{3}\right)=\log _e k \Rightarrow 3 k-2=\sqrt{3}$
$\Rightarrow 9 k^2-12 k+4=3 \Rightarrow 9 k^2-12 k+1=0$
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