If tanh x = operatorname{sech} y = frac{3}{5} | vedclass

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

Question

(D) Given $	anh x = \frac{3}{5}$ $\Rightarrow \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{3}{5}$ $\Rightarrow \frac{e^{2x} - 1}{e^{2x} + 1} = \frac{3}{

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) Given $	anh x = \frac{3}{5}$ $\Rightarrow \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{3}{5}$ $\Rightarrow \frac{e^{2x} - 1}{e^{2x} + 1} = \frac{3}{5}$.$5e^{2x} - 5 = 3e^{2x} + 3$ $\Rightarrow 2e^{2x} = 8$ $\Rightarrow e^{2x} = 4$ $\Rightarrow e^x = 2$.Given $\operatorname{sech} y = \frac{3}{5}$ $\Rightarrow \frac{2}{e^y + e^{-y}} = \frac{3}{5}$ $\Rightarrow \frac{2e^y}{e^{2y} + 1} = \frac{3}{5}$.$10e^y = 3e^{2y} + 3 \Rightarrow 3(e^y)^2 - 10e^y + 3 = 0$.Let $t = e^y$,then $3t^2 - 10t + 3 = 0$ $\Rightarrow (3t - 1)(t - 3) = 0$ $\Rightarrow t = 3$ or $t = \frac{1}{3}$.Since $e^{x+y} = e^x \cdot e^y$,we have $e^{x+y} = 2 	imes 3 = 6$ or $e^{x+y} = 2 	imes \frac{1}{3} = \frac{2}{3}$.Since $e^{x+y}$ is an integer,$e^{x+y} = 6$.

If $\tanh x = \operatorname{sech} y = \frac{3}{5}$ and $e^{x+y}$ is an integer,then $e^{x+y} =$

Similar Questions

Number of real values of $x \in (0, \pi)$ for which $\frac{8}{3\sin x - \sin 3x} + 3\sin^2 x \le 5$ is:

If $\tan \theta \cdot \tan \left(120^{\circ}-\theta\right) \tan \left(120^{\circ}+\theta\right)=\frac{1}{\sqrt{3}}$,then $\theta$ is equal to

The value of $\cos^2 10^o - \cos 10^o \cos 50^o + \cos^2 50^o$ is

If $\frac{\sin^4 x}{2} + \frac{\cos^4 x}{3} = \frac{1}{5}$,then $27 \sec^6 x + 8 \operatorname{cosec}^6 x = $

$\cos ^2 5^{\circ}-\cos ^2 15^{\circ}-\sin ^2 15^{\circ}+\sin ^2 35^{\circ}+\cos 15^{\circ} \sin 15^{\circ}-\cos 5^{\circ} \sin 35^{\circ} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module