Question:

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

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

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 \( \frac{\pi}{4} + \frac{x}{2} \). 
 

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