If x=-frac{1}{2},then sinh ^{-1} x+operatorna | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $x = -\frac{1}{2}$.We know that $\sinh^{-1} x = \ln(x + \sqrt{x^2+1})$ and $\operatorname{cosech}^{-1} x = \ln\left(\frac{1}{x} + \sqrt{\fra

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $x = -\frac{1}{2}$.We know that $\sinh^{-1} x = \ln(x + \sqrt{x^2+1})$ and $\operatorname{cosech}^{-1} x = \ln\left(\frac{1}{x} + \sqrt{\frac{1}{x^2}+1}\right)$.Substituting $x = -\frac{1}{2}$:$\sinh^{-1}\left(-\frac{1}{2}\right) = \ln\left(-\frac{1}{2} + \sqrt{\frac{1}{4}+1}\right) = \ln\left(\frac{\sqrt{5}-1}{2}\right)$.$\operatorname{cosech}^{-1}\left(-\frac{1}{2}\right) = \ln\left(-2 + \sqrt{4+1}\right) = \ln(\sqrt{5}-2)$.Adding these:$\sinh^{-1} x + \operatorname{cosech}^{-1} x = \ln\left(\frac{\sqrt{5}-1}{2}\right) + \ln(\sqrt{5}-2) = \ln\left(\frac{(\sqrt{5}-1)(\sqrt{5}-2)}{2}\right)$.$= \ln\left(\frac{5 - 2\sqrt{5} - \sqrt{5} + 2}{2}\right) = \ln\left(\frac{7-3\sqrt{5}}{2}\right)$.

If $x=-\frac{1}{2}$,then $\sinh ^{-1} x+\operatorname{cosech}^{-1} x=$

Similar Questions

If $a \cos \theta + b \sin \theta = m$ and $a \sin \theta - b \cos \theta = n,$ then ${a^2} + {b^2} = $

If $\sinh x = \frac{12}{5}$,then $\sinh 3x + \cosh 3x = $

$\frac{\sin 1^{\circ}+\sin 2^{\circ}+\ldots+\sin 89^{\circ}}{2(\cos 1^{\circ}+\cos 2^{\circ}+\ldots+\cos 44^{\circ})+1} = $

If $\tan {3^o} + 2\tan {6^o} + 4\tan {12^o} + 8\cot {24^o} = \cot {\theta ^o}$,then:

If $f(x) = \frac{\cos^2 x + \sin^4 x}{\sin^2 x + \cos^4 x}$,for $x \in R$,then $f(2002)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module