જો x=-frac{1}{2} હોય,તો sinh ^{-1} x+operator | vedclass

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(A) આપેલ છે $x = -\frac{1}{2}$.આપણે જાણીએ છીએ કે $\sinh^{-1} x = \ln(x + \sqrt{x^2+1})$ અને $\operatorname{cosech}^{-1} x = \ln\left(\frac{1}{x} + \sq

Vedclass · Accepted Answer

$\log _e\left(\frac{7-3 \sqrt{5}}{2}\right)$

Vedclass · Answer

(A) આપેલ છે $x = -\frac{1}{2}$.આપણે જાણીએ છીએ કે $\sinh^{-1} x = \ln(x + \sqrt{x^2+1})$ અને $\operatorname{cosech}^{-1} x = \ln\left(\frac{1}{x} + \sqrt{\frac{1}{x^2}+1}\right)$.$x = -\frac{1}{2}$ મૂકતા:$\sinh^{-1}\left(-\frac{1}{2}\right) = \ln\left(-\frac{1}{2} + \sqrt{\frac{1}{4}+1}\right) = \ln\left(\frac{\sqrt{5}-1}{2}\right)$.$\operatorname{cosech}^{-1}\left(-\frac{1}{2}\right) = \ln\left(-2 + \sqrt{4+1}\right) = \ln(\sqrt{5}-2)$.બંનેનો સરવાળો કરતા:$\sinh^{-1} x + \operatorname{cosech}^{-1} x = \ln\left(\frac{\sqrt{5}-1}{2}\right) + \ln(\sqrt{5}-2) = \ln\left(\frac{(\sqrt{5}-1)(\sqrt{5}-2)}{2}\right)$.$= \ln\left(\frac{5 - 2\sqrt{5} - \sqrt{5} + 2}{2}\right) = \ln\left(\frac{7-3\sqrt{5}}{2}\right)$.

જો $x=-\frac{1}{2}$ હોય,તો $\sinh ^{-1} x+\operatorname{cosech}^{-1} x=$

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