$\operatorname{sech}^{-1}\left(\frac{1}{\sqrt{2}}\right)+\operatorname{cosech}^{-1}(-1)=$

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Aicharya · Accepted Answer

We have, $\begin{aligned} & \operatorname{sech}^{-1}\left(\frac{1}{\sqrt{2}}ight)+\operatorname{cosec} h^{-1}(-1) \ & =\log _e\left(\frac{1+\sqrt{1-1 / 2}}{\frac{1}{\sqrt{2}}}ight)+\log _e\left(\frac{1-\sqrt{1+1}}{-1}ight) \ & =\log _e\left(\frac{\sqrt{2}+1}{\frac{1}{\sqrt{2}}}ight)+\log _e(\sqrt{2}-1) \ & =\log _e(\sqrt{2}+1)+\log _e(\sqrt{2}-1) \ & =\log _e[(\sqrt{2}+1)(\sqrt{2}-1)] \ \end{aligned}$ $\begin{aligned} & \quad[\because \log m+\log n=\log (m 	imes n)] \ & =\log _e\left((\sqrt{2})^2-1ight)=\log _e(2-1)=\log _e 1 \ & 	herefore \quad \sec h^{-1}\left(\frac{1}{\sqrt{2}}ight)+\operatorname{cosec}^{-1}(-1)=0\end{aligned}$

$\operatorname{sech}^{-1}\left(\frac{1}{\sqrt{2}}\right)+\operatorname{cosech}^{-1}(-1)=$

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