Question

tan inversecosx/1+sinx​

Answer 1

$Given \: \: Question \: \: Is \: \\ \\ \tan {}^{ - 1} ( \frac{ \cos(x) }{1 + \sin(x) } ) \\ \\ \tan {}^{ - 1} ( \frac{ \cos {}^{2} ( \frac{x}{2} ) - \sin {}^{2} ( \frac{x}{2} ) }{( \sin( \frac{x}{2} ) + \cos( \frac{x}{2} ) ) {}^{2} } ) \\ \\ \tan {}^{ - 1} ( \frac{ (\cos( \frac{x}{2} ) - \sin( \frac{x}{2} ) ) \times ( \cos( \frac{x}{2} ) - \sin( \frac{x}{2} )) }{( \sin( \frac{x}{2} ) + \cos( \frac{x}{2} ) ) \times ( \sin( \frac{x}{2} ) + \cos( \frac{x}{2} ) )} ) \\ \\ \tan {}^{ - 1} ( \frac{ \cos( \frac{x}{2} ) - \sin( \frac{x}{2} ) }{ \cos( \frac{x}{2} ) + \sin( \frac{x}{2} )) } ) \\ \\ \tan {}^{ - 1} ( \frac{ \cos( \frac{x}{2} )(1 - \frac{ \sin( \frac{x}{2} ) }{ \cos( \frac{x}{2} ) } }{ \cos( \frac{x}{2} ) (1 + \frac{ \sin( \frac{x}{2} ) }{ \cos( \frac{x}{2} ) } )} ) \\ \\ \tan {}^{ - 1} ( \frac{1 - \tan( \frac{x}{2} ) }{1 + \tan( \frac{x}{2} ) } ) \\ \\ \tan {}^{ - 1} ( \tan( \frac{\pi}{4} - \frac{x}{2} ) ) \\ \\ = \frac{\pi}{4} - \frac{x}{2} \\ \\ therefore \: \: \: \: \: \: \: \: \: \tan( \frac{ \cos(x) }{1 + \sin(x) } ) = \frac{\pi}{4} - \frac{x}{2} \\ \\ Note \: \\ \\ \cos(2x) = \cos {}^{2} (x) - \sin {}^{2} (x) \\ \\ 1 + \sin(x) = ( \sin( \frac{x}{2}) + \cos( \frac{x}{2} ) ) {}^{2}$