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) We know that $\operatorname{sech}^{-1}(x) = \ln\left(\frac{1+\sqrt{1-x^2}}{x}\right)$ and $\operatorname{cosech}^{-1}(x) = \ln\left(\frac{1+\sqrt{

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) We know that $\operatorname{sech}^{-1}(x) = \ln\left(\frac{1+\sqrt{1-x^2}}{x}\right)$ and $\operatorname{cosech}^{-1}(x) = \ln\left(\frac{1+\sqrt{1+x^2}}{x}\right)$.Step $1$: Evaluate $\operatorname{sech}^{-1}\left(\frac{1}{\sqrt{2}}\right)$.$\operatorname{sech}^{-1}\left(\frac{1}{\sqrt{2}}\right) = \ln\left(\frac{1+\sqrt{1-1/2}}{1/\sqrt{2}}\right) = \ln\left(\frac{1+1/\sqrt{2}}{1/\sqrt{2}}\right) = \ln(\sqrt{2}+1)$.Step $2$: Evaluate $\operatorname{cosech}^{-1}(-1)$.$\operatorname{cosech}^{-1}(-1) = \ln\left(\frac{1+\sqrt{1+(-1)^2}}{-1}\right) = \ln\left(\frac{1+\sqrt{2}}{-1}\right) = \ln\left(\frac{1}{\sqrt{2}+1}\right) = \ln((\sqrt{2}+1)^{-1}) = -\ln(\sqrt{2}+1)$.Step $3$: Add the results.$\operatorname{sech}^{-1}\left(\frac{1}{\sqrt{2}}\right)+\operatorname{cosech}^{-1}(-1) = \ln(\sqrt{2}+1) - \ln(\sqrt{2}+1) = 0$.

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

Similar Questions

The value of $\sin \left(2 \sin ^{-1} 0.8\right)$ is equal to

$\tan ^{-1}(-\sqrt{3})-\sec ^{-1}(-2)=$ . . . . . . .

Find the principal value of $\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$.

Solve $\sin(\tan^{-1} x)$,where $|x| < 1$,is equal to:

Find the value of $\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module