Question

Simplify:
\( \tan^{-1} \left( \frac{\cos x}{1 - \sin x} \right) \)

Answer 1

Multiply numerator and denominator by \( 1 + \sin x \):
\[ \tan^{-1} \left( \frac{\cos x}{1 - \sin x} \cdot \frac{1 + \sin x}{1 + \sin x} \right) = \tan^{-1} \left( \frac{\cos x (1 + \sin x)}{(1 - \sin x)(1 + \sin x)} \right) \] 
Denominator becomes:
\[ (1 - \sin x)(1 + \sin x) = 1 - \sin^2 x = \cos^2 x \] 
Numerator becomes:
\[ \cos x (1 + \sin x) \] 
So the expression becomes:
\[ \tan^{-1} \left( \frac{\cos x (1 + \sin x)}{\cos^2 x} \right) \] 
Simplify:
\[ \tan^{-1} \left( \frac{1 + \sin x}{\cos x} \right) \] 
Now, use identity:
\[ \frac{1 + \sin x}{\cos x} = \tan \left( \frac{\pi}{4} + \frac{x}{2} \right) \] 
Therefore,
\[ \tan^{-1} \left( \frac{\cos x}{1 - \sin x} \right) = \frac{\pi}{4} + \frac{x}{2} \]