Question

prove - tan(π/4 +A ) = 1 + sin 2A/cos 2A

Answer 1

$\tan( \frac{\pi}{4} + \alpha ) = \frac{ \tan( \frac{\pi}{4} ) + \tan( \alpha ) }{1 - \tan( \frac{\pi}{4} ) \tan( \alpha ) } \\ = \frac{1 + \tan( \alpha ) }{1 - \tan( \alpha ) } = \frac{1 + \frac{ \sin( \alpha ) }{ \cos( \alpha ) } }{1 - \frac{ \sin( \alpha ) }{ \cos( \alpha ) } } \\ = \frac{ \cos( \alpha ) + \sin( \alpha ) }{ \cos( \alpha ) - \sin( \alpha ) } \\ = \frac{ { (\cos( \alpha ) + \sin( \alpha ))}^{2} }{( \cos( \alpha ) - \sin( \alpha )( \cos( \alpha ) + \sin( \alpha ))} \\ = \frac{{ \cos}^{2} \alpha + { \sin}^{2} \alpha + 2 \cos( \alpha ) \sin( \alpha ) }{{ \cos}^{2} \alpha - { \sin}^{2} \alpha} \\ = \frac{1 + 2 \cos( \alpha ) \sin( \alpha ) }{{ \cos}^{2} \alpha - { \sin}^{2} \alpha} \\ = \frac{1 + \sin(2 \alpha ) }{ \cos(2 \alpha ) }$