Assertion (A): operatorname{cosech}^{-1}(3) = | vedclass

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(A) For Reason $(R)$,let $p = e^{\operatorname{cosech}^{-1} x}$.Then $\operatorname{cosech}^{-1} x = \ln p$,which implies $x = \operatorname{cosech}(\

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$(A)$ is true,$(R)$ is true and $(R)$ is the correct explanation for $(A)$

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(A) For Reason $(R)$,let $p = e^{\operatorname{cosech}^{-1} x}$.Then $\operatorname{cosech}^{-1} x = \ln p$,which implies $x = \operatorname{cosech}(\ln p) = \frac{e^{\ln p} - e^{-\ln p}}{2} = \frac{p - 1/p}{2} = \frac{p^2 - 1}{2p}$.Rearranging gives $2px = p^2 - 1$,or $p^2 - 2px - 1 = 0$.Wait,the given equation is $x p^2 - 2p - x = 0$. Let us check if $p = e^{\operatorname{cosech}^{-1} x}$ satisfies this.We know $\operatorname{cosech}^{-1} x = \ln \left(\frac{1 + \sqrt{1+x^2}}{x}\right)$.So $p = \frac{1 + \sqrt{1+x^2}}{x}$.Then $xp^2 - 2p - x = x \left(\frac{1 + \sqrt{1+x^2}}{x}\right)^2 - 2 \left(\frac{1 + \sqrt{1+x^2}}{x}\right) - x = \frac{1 + 1 + x^2 + 2\sqrt{1+x^2}}{x} - \frac{2 + 2\sqrt{1+x^2}}{x} - x = \frac{2 + x^2 + 2\sqrt{1+x^2} - 2 - 2\sqrt{1+x^2}}{x} - x = \frac{x^2}{x} - x = x - x = 0$.Thus,Reason $(R)$ is true.For Assertion $(A)$,$\operatorname{cosech}^{-1}(3) = \ln \left(\frac{1 + \sqrt{1+3^2}}{3}\right) = \ln \left(\frac{1 + \sqrt{10}}{3}\right)$.Thus,Assertion $(A)$ is true and $(R)$ explains $(A)$.

Assertion $(A): \operatorname{cosech}^{-1}(3) = \log \left(\frac{1+\sqrt{10}}{3}\right)$
Reason $(R): e^{\operatorname{cosech}^{-1} x}$ is a root of the quadratic equation $x p^2 - 2p - x = 0$
The correct option among the following is

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Assertion $(A): \operatorname{cosech}^{-1}(3) = \log \left(\frac{1+\sqrt{10}}{3}\right)$Reason $(R): e^{\operatorname{cosech}^{-1} x}$ is a root of the quadratic equation $x p^2 - 2p - x = 0$The correct option among the following is