Question

If \(\cos \alpha = \sec h \beta\), then \(\beta =\)

• A. \(\log(\sec \alpha + \tan \alpha)\)
• B. \(\log(\sec \alpha - \tan \alpha)\)
• C. \(\log(\sin \alpha + \cos \alpha)\)
• D. \(\log(\cos \alpha + \cot \alpha)\)

Hint:

Use definitions and quadratic formula to find relationships between hyperbolic and trigonometric functions.

Answer 1

The Correct Option is A

Step 1: Use definition of hyperbolic secant
\[ \sech \beta = \frac{2}{e^\beta + e^{-\beta}} = \cos \alpha \] Step 2: Rearrange for \(e^\beta\)
\[ \cos \alpha = \frac{2}{e^\beta + e^{-\beta}} \implies e^\beta + e^{-\beta} = \frac{2}{\cos \alpha} \] Step 3: Let \(x = e^\beta\)
\[ x + \frac{1}{x} = \frac{2}{\cos \alpha} \implies x^2 - \frac{2}{\cos \alpha} x + 1 = 0 \] Step 4: Solve quadratic for \(x\)
\[ x = \frac{\frac{2}{\cos \alpha} \pm \sqrt{\left(\frac{2}{\cos \alpha}\right)^2 - 4}}{2} = \frac{1}{\cos \alpha} \pm \frac{\tan \alpha}{\cos \alpha} \] Step 5: Simplify
\[ x = \sec \alpha \pm \tan \alpha \] Taking positive root, \[ e^\beta = \sec \alpha + \tan \alpha \] Step 6: Take logarithm
\[ \beta = \log (\sec \alpha + \tan \alpha) \]