Question

Sech\(^{-1}(\sin \alpha)\) is equal to:

• A. \( \log(\sin \alpha + \sqrt{\sin^2 \alpha - 1}) \)
• B. \( \log(\tan \alpha + 1) \)
• C. \( \log\left(\csc \alpha + \cot \alpha\right) \)
• D. \( \log\left(\frac{1 + \tan \alpha}{2 \sin \alpha}\right) \)

Hint:

Remember inverse hyperbolic identities: \(\sech^{-1}(x) = \log\left(\frac{1 + \sqrt{1 - x^2}}{x}\right)\) for \(0<x \leq 1\). Expressing hyperbolic and trigonometric inverses in logarithmic form is a common trick.

Answer 1

The Correct Option is C

Let \( y = \sech^{-1}(\sin \alpha) \). By definition, \[ \sech(y) = \sin \alpha \] \[ \frac{1}{\cosh y} = \sin \alpha \] \[ \cosh y = \frac{1}{\sin \alpha} \] Recall that \(\cosh y = \frac{e^y + e^{-y}}{2}\): \[ \frac{e^y + e^{-y}}{2} = \frac{1}{\sin \alpha} \] \[ e^y + e^{-y} = \frac{2}{\sin \alpha} \] Let \( z = e^y \): \[ z + \frac{1}{z} = \frac{2}{\sin \alpha} \] \[ z^2 - \frac{2}{\sin \alpha}z + 1 = 0 \] Solving this quadratic in \(z\): \[ z = \frac{\frac{2}{\sin \alpha} \pm \sqrt{\left(\frac{2}{\sin \alpha}\right)^2 - 4}}{2} \] \[ = \frac{1}{\sin \alpha} \pm \frac{\sqrt{1 - \sin^2 \alpha}}{\sin \alpha} \] \[ = \frac{1}{\sin \alpha} \pm \frac{\cos \alpha}{\sin \alpha} \] \[ = \csc \alpha \pm \cot \alpha \] Since \(y>0\), take the positive root: \[ e^y = \csc \alpha + \cot \alpha \] \[ y = \log(\csc \alpha + \cot \alpha) \]