Question

Express \( \tan^{-1} \left( \frac{\cos x}{1 - \sin x} \right) \), where \( -\frac{\pi}{2}<x<\frac{\pi}{2} \), in the simplest form.

Hint:

For expressions involving \( \tan^{-1} \), rewrite in terms of trigonometric identities to simplify.

Answer 1

Step 1: {Simplify the expression inside \( \tan^{-1} \)}
The given expression is: \[ \tan^{-1} \left( \frac{\cos x}{1 - \sin x} \right). \] Using trigonometric identities, rewrite: \[ 1 - \sin x = (\cos^2\frac{x}{2} + \sin^2\frac{x}{2}) - 2\sin\frac{x}{2}\cos\frac{x}{2} = (\cos\frac{x}{2} - \sin\frac{x}{2})^2. \] Step 2: {Transform into a single tangent function}
Substituting \( 1 - \sin x \) and \( \cos x = (\cos^2\frac{x}{2} - \sin^2\frac{x}{2}) \): \[ \tan^{-1} \left( \frac{\cos x}{1 - \sin x} \right) = \tan^{-1} \left[ \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) \right]. \] Step 3: {Simplify using \( \tan^{-1} \tan y = y \)}
Since \( -\frac{\pi}{2}<x<\frac{\pi}{2} \), we simplify: \[ \tan^{-1} \left[ \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) \right] = \frac{\pi}{4} + \frac{x}{2}. \] Conclusion: The simplest form is \( \frac{\pi}{4} + \frac{x}{2} \).