Question

Simplify
Inverse trigonometry

Answer 1

Step-by-step explanation:

We have,

$\tan^{ - 1} \bigg \{\frac{ \cos(x) }{1 - \sin(x) } \bigg \}$

$= \tan^{ - 1} \bigg \{\frac{ \cos^{2} ( \frac{x}{2} ) - \sin ^{2} ( \frac{x}{2} ) }{1 - 2\sin( \frac{x}{2} ) \cos( \frac{x}{2} ) } \bigg \}$

$= \tan^{ - 1} \bigg [ \frac{ \{\cos ( \frac{x}{2} ) - \sin ( \frac{x}{2} ) \} \{ \cos( \frac{x}{2} ) + \sin( \frac{x}{2} ) \} }{ \cos^{2} ( \frac{x}{2} ) + \sin ^{2} ( \frac{x}{2} ) - 2\sin( \frac{x}{2} ) \cos( \frac{x}{2} ) } \bigg]$

$= \tan^{ - 1} \bigg [ \frac{ \{\cos ( \frac{x}{2} ) - \sin ( \frac{x}{2} ) \} \{ \cos( \frac{x}{2} ) + \sin( \frac{x}{2} ) \} }{ \{ \cos ( \frac{x}{2} ) + \sin ( \frac{x}{2} ) \}^{2} } \bigg]$

$= \tan^{ - 1} \bigg [ \frac{ \{\cos ( \frac{x}{2} ) - \sin ( \frac{x}{2} ) \} \{ \cos( \frac{x}{2} ) + \sin( \frac{x}{2} ) \} }{ \{ \cos ( \frac{x}{2} ) + \sin ( \frac{x}{2} ) \} \{ \cos ( \frac{x}{2} ) + \sin ( \frac{x}{2} ) \}} \bigg]$

$= \tan^{ - 1} \bigg [ \frac{ \cos ( \frac{x}{2} ) - \sin ( \frac{x}{2} ) }{ \cos ( \frac{x}{2} ) + \sin ( \frac{x}{2} ) } \bigg]$

$= \tan^{ - 1} \bigg [ \frac{ \frac{ \cos ( \frac{x}{2} ) - \sin ( \frac{x}{2} ) }{ \cos( \frac{x}{2} ) } }{ \frac{ \cos ( \frac{x}{2} ) + \sin ( \frac{x}{2} )}{ \cos( \frac{x}{2} ) } } \bigg]$

$= \tan^{ - 1} \bigg [ \frac{1 - \tan ( \frac{x}{2} ) }{1 + \tan( \frac{x}{2} ) } \bigg]$

$= \tan^{ - 1} \bigg [ \frac{ \tan( \frac{\pi}{4} ) - \tan ( \frac{x}{2} ) }{1 + \tan( \frac{\pi}{2} ) \tan( \frac{x}{2} ) } \bigg]$

$= \tan^{ - 1} \bigg \{ \tan \bigg( \frac{\pi}{4} - \frac{x}{2} \bigg ) \bigg \}$

$= \frac{\pi}{4} - \frac{x}{2}$