यदि 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) दिया गया है $	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} = \fr

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) दिया गया है $	anh x = \frac{3}{5}$ $\Rightarrow \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{3}{5}$ $\Rightarrow \frac{e^{2x} - 1}{e^{2x} + 1} = \frac{3}{5}$.$5e^{2x} - 5 = 3e^{2x} + 3$ $\Rightarrow 2e^{2x} = 8$ $\Rightarrow e^{2x} = 4$ $\Rightarrow e^x = 2$.दिया गया है $\operatorname{sech} y = \frac{3}{5}$ $\Rightarrow \frac{2}{e^y + e^{-y}} = \frac{3}{5}$ $\Rightarrow \frac{2e^y}{e^{2y} + 1} = \frac{3}{5}$.$10e^y = 3e^{2y} + 3 \Rightarrow 3(e^y)^2 - 10e^y + 3 = 0$.माना $t = e^y$,तब $3t^2 - 10t + 3 = 0$ $\Rightarrow (3t - 1)(t - 3) = 0$ $\Rightarrow t = 3$ या $t = \frac{1}{3}$.चूंकि $e^{x+y} = e^x \cdot e^y$,इसलिए $e^{x+y} = 2 	imes 3 = 6$ या $e^{x+y} = 2 	imes \frac{1}{3} = \frac{2}{3}$.चूंकि $e^{x+y}$ एक पूर्णांक है,इसलिए $e^{x+y} = 6$.

यदि $\tanh x = \operatorname{sech} y = \frac{3}{5}$ और $e^{x+y}$ एक पूर्णांक है,तो $e^{x+y} =$

Similar Questions

$\cos ^4 \frac{\pi}{12} + \cos ^4 \frac{5 \pi}{12} + \cos ^4 \frac{7 \pi}{12} + \cos ^4 \frac{11 \pi}{12} = $

मान ज्ञात कीजिए: $\sqrt{\sin ^4 x+4 \cos ^2 x}-\sqrt{\cos ^4 x+4 \sin ^2 x}$

$\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{14 \pi}{15} = $

$6(\sin^6 \theta + \cos^6 \theta) - 9(\sin^4 \theta + \cos^4 \theta) + 4$ का मान है

$\sin \frac{\pi}{16} \sin \frac{3\pi}{16} \sin \frac{5\pi}{16} \sin \frac{7\pi}{16}$ का मान क्या है?

Vedclass Test Series

Exam Paper Generator

Online Exam Module