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

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

Question

(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

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(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।$\operatorname{sech}^{-1}\left(\frac{1}{\sqrt{2}}\right)+\operatorname{cosech}^{-1}(-1) = \ln(\sqrt{2}+1) - \ln(\sqrt{2}+1) = 0$.

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

Similar Questions

$\tan ^{-1}(-\sqrt{3})-\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)+\cos ^{-1}\left(\frac{-1}{2}\right)$ का मान ज्ञात कीजिए।

$\sin^{-1}\left(\sin \frac{5\pi}{3}\right)$ का मान . . . . . . के बराबर है।

$\sin ^{-1}\left(\frac{1}{2}\right)+\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)+\cot ^{-1}\left(-\frac{1}{\sqrt{3}}\right)=$

$\cot \left[ \cos^{-1} \left( \frac{7}{25} \right) \right] = $

$\tan ^{-1} 2 x+\tan ^{-1} 3 x=\frac{\pi}{4}$ को हल करें।

Vedclass Test Series

Exam Paper Generator

Online Exam Module