यदि operatorname{sech}^{-1} x = log 2 और oper | vedclass

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(B) दिया गया है $\operatorname{sech}^{-1} x = \log 2$।परिभाषा के अनुसार, $\operatorname{sech}^{-1} x = \cosh^{-1} (\frac{1}{x}) = \log 2$।अतः, $\cosh^

Vedclass · Accepted Answer

$\frac{1}{20}$

Vedclass · Answer

(B) दिया गया है $\operatorname{sech}^{-1} x = \log 2$।परिभाषा के अनुसार, $\operatorname{sech}^{-1} x = \cosh^{-1} (\frac{1}{x}) = \log 2$।अतः, $\cosh^{-1} (\frac{1}{x}) = \log 2$, जिसका अर्थ है $\frac{1}{x} = \cosh(\log 2)$।चूंकि $\cosh(	heta) = \frac{e^{	heta} + e^{-	heta}}{2}$, इसलिए $\frac{1}{x} = \frac{e^{\log 2} + e^{-\log 2}}{2} = \frac{2 + \frac{1}{2}}{2} = \frac{5/2}{2} = \frac{5}{4}$।अतः, $x = \frac{4}{5}$।दिया गया है $\operatorname{cosech}^{-1} y = -\log 3$।परिभाषा के अनुसार, $\operatorname{cosech}^{-1} y = \sinh^{-1} (\frac{1}{y}) = -\log 3$।अतः, $\frac{1}{y} = \sinh(-\log 3) = -\sinh(\log 3)$।चूंकि $\sinh(	heta) = \frac{e^{	heta} - e^{-	heta}}{2}$, इसलिए $\frac{1}{y} = -(\frac{e^{\log 3} - e^{-\log 3}}{2}) = -(\frac{3 - 1/3}{2}) = -(\frac{8/3}{2}) = -\frac{4}{3}$।अतः, $y = -\frac{3}{4}$।इसलिए, $x + y = \frac{4}{5} - \frac{3}{4} = \frac{16 - 15}{20} = \frac{1}{20}$।

यदि $\operatorname{sech}^{-1} x = \log 2$ और $\operatorname{cosech}^{-1} y = -\log 3$ है, तो $(x + y) = $

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