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

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(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 $\beta =$

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