Question

Please solve as fast as you can :-
Simply :- tan
$\tan^{ - 1}( \ \cos(x) \div 1 - \sin(x))$
​

Answer 1

$Answer \: \\ \\ \tan {}^{ - 1} ( \frac{ \cos(x) }{1 - \sin(x) } ) \\ \\ replace \: \: \: \cos(x) \: \: by \: \: \: \cos {}^{2} ( \frac{x}{2} ) - \sin {}^{2} ( \frac{x}{2} ) \\ and \: \: \: \\ 1 - \sin(x) \: \: by \: \: \: ( \cos( \frac{x}{2}) - \sin( \frac{x}{2} ) ) {}^{2} \\ \\ \\ \tan {}^{ - 1} ( \frac{ \cos {}^{2} ( \frac{x}{2} ) - \sin {}^{2} ( \frac{x}{2} ) }{( \cos( \frac{x}{2} ) - \sin( \frac{x}{2} )) {}^{2} } ) \\ \\ \\ \tan {}^{ - 1} ( \frac{( \cos( \frac{x}{2} ) - \sin( \frac{x}{2} ) ) \times( \cos( \frac{x}{2} ) + \sin( \frac{x}{2} )) }{( \cos( \frac{x}{2} ) - \sin( \frac{x}{2} )) \times ( \cos( \frac{x}{2} ) - \sin( \frac{x}{2} )) } ) \\ \\ \\ \tan {}^{ - 1} ( \frac{ \cos( \frac{x}{2} ) + \sin( \frac{x}{2} ) }{ \cos( \frac{x}{2} ) - \sin( \frac{x}{2} ) } ) \\ \\ \\ \tan {}^{ - 1} ( \frac{ \cos( \frac{x}{2} )(1 + \frac{ \sin( \frac{x}{2} ) }{ \cos( \frac{x}{2} ) }) }{ \cos( \frac{x}{2} )(1 - \frac{ \sin( \frac{x}{2} ) }{ \cos( \frac{x}{2} ) }) } ) \\ \\ \tan {}^{ - 1} ( \frac{1 + \tan( \frac{x}{2} ) }{1 - \tan( \frac{x}{2} ) } ) \\ \\ \tan {}^{ - 1} ( \tan ( \frac{\pi}{4} + \frac{x}{2} ) ) \\ \\ \frac{\pi}{4} + \frac{x}{2} \\ \\ therefore \: \: \: \: \: \: \\ \\ \tan( \frac{ \cos(x) }{1 - \sin(x) } ) = \frac{\pi}{4} + \frac{x}{2} \\ \\ \\ NOTE \: \\ \\ \cos(2 \alpha ) = \cos {}^{2} ( \alpha ) - \sin {}^{2} ( \alpha )$