Question

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

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $-\frac{1}{3}$
• D. $-\frac{2}{3}$

Answer 1

Answer: (D)

$\because \cos h^{2} x-\sin h^{2} x=1$
$\Rightarrow \cos h^{2} x-\cos ^{2} \theta=1 [\because \sin\,h x=\cos \theta] $
$\Rightarrow \cos ^{2} \theta=\cos\,h ^{2} x-1 $
$=\frac{14}{9}-1=\frac{5}{9} $
So, $\sin ^{2} \theta=1-\cos ^{2} \theta=1-\frac{5}{9}=\frac{4}{9} $
$ \because -\pi<\theta<-\frac{\pi}{2} $

$ \therefore \sin \theta=-\frac{2}{3}$