If cos alpha = operatorname{sech} beta,then b | vedclass

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

Question

(A) Given that $\cos \alpha = \operatorname{sech} \beta$. Since $\operatorname{sech} \beta = \frac{1}{\cosh \beta}$,we have $\cosh \beta = \frac{1}{\c

Vedclass · Accepted Answer

$\log (\sec \alpha + 	an \alpha)$

Vedclass · Answer

(A) Given that $\cos \alpha = \operatorname{sech} \beta$. Since $\operatorname{sech} \beta = \frac{1}{\cosh \beta}$,we have $\cosh \beta = \frac{1}{\cos \alpha} = \sec \alpha$. Using the definition of hyperbolic cosine,$\cosh \beta = \frac{e^{\beta} + e^{-\beta}}{2} = \sec \alpha$. Let $e^{\beta} = x$. Then $x + \frac{1}{x} = 2 \sec \alpha$,which implies $x^2 - (2 \sec \alpha) x + 1 = 0$. Solving for $x$ using the quadratic formula: $x = \frac{2 \sec \alpha \pm \sqrt{4 \sec^2 \alpha - 4}}{2} = \sec \alpha \pm \sqrt{\sec^2 \alpha - 1} = \sec \alpha \pm 	an \alpha$. Since $e^{\beta} = \sec \alpha + 	an \alpha$ (taking the positive root for real $\beta$),we get $\beta = \log (\sec \alpha + 	an \alpha)$.

If $\cos \alpha = \operatorname{sech} \beta$,then $\beta =$

Similar Questions

If $\sec(\theta+\alpha)$,$\sec\theta$,and $\sec(\theta-\alpha)$ are in arithmetic progression,then the value of $\cos\theta \cdot \sec\frac{\alpha}{2}$ is:

Which of the following statements is/are correct for $0 < \theta < \frac{\pi}{2}$?

If $A$ is in the third quadrant and $\tan A = \frac{\sqrt{7}}{3}$,then $18 - 16 \sin^2 \frac{A}{2} - 32 \sin \frac{A}{2} \sin \frac{5A}{2} = $

If $\frac{\sqrt{2} \sin \alpha}{\sqrt{1+\cos 2 \alpha}}=\frac{1}{7}$ and $\sqrt{\frac{1-\cos 2 \beta}{2}}=\frac{1}{\sqrt{10}}$ where $\alpha, \beta \in (0, \frac{\pi}{2})$,then $\tan (\alpha+2 \beta)$ is equal to

If $f(x) = \frac{\cos^2 x + \sin^4 x}{\sin^2 x + \cos^4 x}$,for $x \in R$,then $f(2002)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module