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

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

Question

(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 $\sin \theta =$

Similar Questions

Let $S$ be the set of all $\alpha \in \mathbb{R}$ such that the equation $\cos 2x + \alpha \sin x = 2\alpha - 7$ has a solution. Then $S$ is equal to

If $\cos x+\cos y=p$ and $\sin x+\sin y=q$,then $\cos \left(\frac{x-y}{2}\right) = $

In a right-angled triangle,the hypotenuse is $2 \sqrt{2}$ times the perpendicular drawn from the opposite vertex. Then the other acute angles of the triangle are

Let $S = \{\theta \in [0, 2\pi] : 8^{2 \sin^2 \theta} + 8^{2 \cos^2 \theta} = 16\}$. Then $n(S) + \sum_{\theta \in S} \left(\sec \left(\frac{\pi}{4} + 2\theta\right) \operatorname{cosec} \left(\frac{\pi}{4} + 2\theta\right)\right)$ is equal to.

If $a \cos \theta + b \sin \theta = m$ and $a \sin \theta - b \cos \theta = n,$ then ${a^2} + {b^2} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module