If operatorname{Tanh}^{-1} x = operatorname{C | vedclass

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

Question

(A) Given $\operatorname{Tanh}^{-1} x = \log \sqrt{5}$.Using the definition $\operatorname{Tanh}^{-1} x = \frac{1}{2} \log \left( \frac{1+x}{1-x} \rig

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(A) Given $\operatorname{Tanh}^{-1} x = \log \sqrt{5}$.Using the definition $\operatorname{Tanh}^{-1} x = \frac{1}{2} \log \left( \frac{1+x}{1-x} \right)$,we have $\frac{1}{2} \log \left( \frac{1+x}{1-x} \right) = \log \sqrt{5} = \frac{1}{2} \log 5$.Thus,$\frac{1+x}{1-x} = 5 \implies 1+x = 5-5x \implies 6x = 4 \implies x = \frac{2}{3}$.Given $\operatorname{Coth}^{-1} y = \log \sqrt{5}$.Using the definition $\operatorname{Coth}^{-1} y = \frac{1}{2} \log \left( \frac{y+1}{y-1} \right)$,we have $\frac{1}{2} \log \left( \frac{y+1}{y-1} \right) = \frac{1}{2} \log 5$.Thus,$\frac{y+1}{y-1} = 5 \implies y+1 = 5y-5 \implies 4y = 6 \implies y = \frac{3}{2}$.Now,$xy = \left( \frac{2}{3} \right) \left( \frac{3}{2} \right) = 1$.Therefore,$\operatorname{Tan}^{-1}(xy) = \operatorname{Tan}^{-1}(1) = \frac{\pi}{4}$.

If $\operatorname{Tanh}^{-1} x = \operatorname{Coth}^{-1} y = \log \sqrt{5}$,then $\operatorname{Tan}^{-1}(xy) = $

Similar Questions

The value of $\sin \left[2 \cos ^{-1} \frac{\sqrt{5}}{3}\right]$ is

The interval for which ${\sin ^{ - 1}}\sqrt x + {\cos ^{ - 1}}\sqrt x = \frac{\pi }{2}$ holds is:

The principal value of $\tan^{-1} \left( \cot \frac{43\pi}{4} \right)$ is

The principal value of $\cos^{-1}[\cos(-680^{\circ})]$ is equal to: . . . . . . .

If $\theta = \sin^{-1}[\sin(-600^\circ)]$,then one of the possible values of $\theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module