If cosh x = frac{sqrt{14}}{3},sinh x = cos th | vedclass

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

Question

(B) Given $\cosh x = \frac{\sqrt{14}}{3}$.Using the identity $\sinh^2 x = \cosh^2 x - 1$,we get:$\sinh^2 x = \left(\frac{\sqrt{14}}{3}\right)^2 - 1 =

Vedclass · Accepted Answer

$\frac{2}{3}$

Vedclass · Answer

(B) Given $\cosh x = \frac{\sqrt{14}}{3}$.Using the identity $\sinh^2 x = \cosh^2 x - 1$,we get:$\sinh^2 x = \left(\frac{\sqrt{14}}{3}ight)^2 - 1 = \frac{14}{9} - 1 = \frac{5}{9}$.Since $\sinh x = \cos 	heta$,we have $\cos 	heta = \frac{\sqrt{5}}{3}$ (taking the positive root as $\sinh x$ is positive for $x > 0$ and $\cos 	heta$ is positive in the given range).Now,$\sin^2 	heta = 1 - \cos^2 	heta = 1 - \frac{5}{9} = \frac{4}{9}$.Thus,$\sin 	heta = \pm \frac{2}{3}$.Given $-\pi In the second quadrant,$\sin 	heta$ is positive.However,checking the range $-\pi Wait,re-evaluating the quadrant: $-\pi  0$.Therefore,$\sin 	heta = \frac{2}{3}$.

If $\cosh x = \frac{\sqrt{14}}{3}$,$\sinh x = \cos \theta$ and $-\pi < \theta < -\frac{\pi}{2}$,then $\sin \theta =$

Similar Questions

If $\tan \theta = - \frac{1}{\sqrt{10}}$ and $\theta$ lies in the fourth quadrant,then $\cos \theta = $

If $\sin (A + B) = 1$ and $\cos (A - B) = \frac{\sqrt{3}}{2}$,then the smallest positive values of $A$ and $B$ are

$\sin 120^{\circ} \cos 150^{\circ} - \cos 240^{\circ} \sin 330^{\circ}$ is equal to :

Which of the following equations has no solution?

If $x = \tan 15^{\circ}$,$y = \operatorname{cosec} 75^{\circ}$ and $z = 4 \sin 18^{\circ}$,then :

Vedclass Test Series

Exam Paper Generator

Online Exam Module