If theta is an acute angle,cosh x = K and sin | vedclass

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(C) Given $\cosh x = K$ and $\sinh x = 	an 	heta$. We know the identity for hyperbolic functions: $\cosh^2 x - \sinh^2 x = 1$. Substituting the give

Vedclass · Accepted Answer

$\frac{\sqrt{K^2-1}}{K}$

Vedclass · Answer

(C) Given $\cosh x = K$ and $\sinh x = 	an 	heta$. We know the identity for hyperbolic functions: $\cosh^2 x - \sinh^2 x = 1$. Substituting the given values: $K^2 - \sinh^2 x = 1$. Therefore,$\sinh^2 x = K^2 - 1$,which implies $\sinh x = \sqrt{K^2 - 1}$. Since $\sinh x = 	an 	heta$,we have $	an 	heta = \sqrt{K^2 - 1} = \frac{\sqrt{K^2 - 1}}{1}$. In a right-angled triangle,$	an 	heta = \frac{	ext{Perpendicular}}{	ext{Base}} = \frac{\sqrt{K^2 - 1}}{1}$. The Hypotenuse $H = \sqrt{(\sqrt{K^2 - 1})^2 + 1^2} = \sqrt{K^2 - 1 + 1} = \sqrt{K^2} = K$. Thus,$\sin 	heta = \frac{	ext{Perpendicular}}{	ext{Hypotenuse}} = \frac{\sqrt{K^2 - 1}}{K}$.

If $\theta$ is an acute angle,$\cosh x = K$ and $\sinh x = \tan \theta$,then $\sin \theta =$

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