operatorname{sech}^{-1}left(frac{1}{sqrt{2}}r | vedclass

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(A) हम जानते हैं कि $\operatorname{sech}^{-1}(x) = \ln\left(\frac{1+\sqrt{1-x^2}}{x}\right)$ और $\operatorname{cosech}^{-1}(x) = \ln\left(\frac{1+\sqr

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(A) हम जानते हैं कि $\operatorname{sech}^{-1}(x) = \ln\left(\frac{1+\sqrt{1-x^2}}{x}\right)$ और $\operatorname{cosech}^{-1}(x) = \ln\left(\frac{1+\sqrt{1+x^2}}{x}\right)$.चरण $1$: $\operatorname{sech}^{-1}\left(\frac{1}{\sqrt{2}}\right)$ का मान ज्ञात करें।$\operatorname{sech}^{-1}\left(\frac{1}{\sqrt{2}}\right) = \ln\left(\frac{1+\sqrt{1-1/2}}{1/\sqrt{2}}\right) = \ln\left(\frac{1+1/\sqrt{2}}{1/\sqrt{2}}\right) = \ln(\sqrt{2}+1)$.चरण $2$: $\operatorname{cosech}^{-1}(-1)$ का मान ज्ञात करें।$\operatorname{cosech}^{-1}(-1) = \ln\left(\frac{1+\sqrt{1+(-1)^2}}{-1}\right) = \ln\left(\frac{1+\sqrt{2}}{-1}\right) = \ln\left(\frac{1}{\sqrt{2}+1}\right) = \ln((\sqrt{2}+1)^{-1}) = -\ln(\sqrt{2}+1)$.चरण $3$: परिणामों को जोड़ें।$\operatorname{sech}^{-1}\left(\frac{1}{\sqrt{2}}\right)+\operatorname{cosech}^{-1}(-1) = \ln(\sqrt{2}+1) - \ln(\sqrt{2}+1) = 0$.

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

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