If operatorname{sech}^{-1}left(frac{1}{2}righ | vedclass

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(C) We know that $\operatorname{sech}^{-1}(x) = \log_e \left( \frac{1+\sqrt{1-x^2}}{x} \right)$ and $\operatorname{cosech}^{-1}(x) = \log_e \left( \fr

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$9 k^2-12 k+1=0$

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(C) We know that $\operatorname{sech}^{-1}(x) = \log_e \left( \frac{1+\sqrt{1-x^2}}{x} \right)$ and $\operatorname{cosech}^{-1}(x) = \log_e \left( \frac{1}{x} + \sqrt{\frac{1}{x^2}+1} \right)$.Given equation: $\operatorname{sech}^{-1}\left(\frac{1}{2}\right)-\operatorname{cosech}^{-1}\left(\frac{3}{4}\right)=\log _e k$.First, calculate $\operatorname{sech}^{-1}\left(\frac{1}{2}\right) = \log_e \left( \frac{1+\sqrt{1-(1/2)^2}}{1/2} \right) = \log_e \left( 2(1+\sqrt{3/4}) \right) = \log_e (2+\sqrt{3})$.Next, calculate $\operatorname{cosech}^{-1}\left(\frac{3}{4}\right) = \log_e \left( \frac{4}{3} + \sqrt{\frac{16}{9}+1} \right) = \log_e \left( \frac{4}{3} + \sqrt{\frac{25}{9}} \right) = \log_e \left( \frac{4}{3} + \frac{5}{3} \right) = \log_e(3)$.Substituting these into the equation: $\log_e(2+\sqrt{3}) - \log_e(3) = \log_e k$.$\log_e \left( \frac{2+\sqrt{3}}{3} \right) = \log_e k$.So, $k = \frac{2+\sqrt{3}}{3}$, which implies $3k - 2 = \sqrt{3}$.Squaring both sides: $(3k-2)^2 = 3$.$9k^2 - 12k + 4 = 3$.$9k^2 - 12k + 1 = 0$.

If $\operatorname{sech}^{-1}\left(\frac{1}{2}\right)-\operatorname{cosech}^{-1}\left(\frac{3}{4}\right)=\log _e k$, then

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