Question

hi guys, please solve this question fast :::::

prove that ,
$\tan( \alpha ) + 2 \tan(2 \alpha ) + 4 \cot(4 \alpha ) = \cot( \alpha )$

Answer 1

We know that,
Steps:
1)
$\cot( \alpha ) - \tan( \alpha ) \\ = \frac{ \cos( \alpha ) }{ \sin( \alpha ) } - \frac{ \sin( \alpha ) }{ \cos( \alpha ) } \\ = \frac{ {( \cos( \alpha ) ) }^{2} - { (\sin( \alpha ) )}^{2} }{ \sin( \alpha ) \cos( \alpha ) } \times \frac{2}{2} \\ = \frac{2 \cos(2 \alpha ) }{ \sin(2 \alpha ) } = 2 \cot(2 \alpha )$
Similarly,
2)
$2 \cot(4 \alpha ) = \cot(2 \alpha ) - \tan(2 \alpha ) \\ - - - (1)$
Now, 3)
$\tan( \alpha ) + 2 \tan(2 \alpha ) + 4 \cot(4 \alpha ) \\ = \tan( \alpha ) + 2 \tan(2 \alpha ) + \\ \: \: \: \: 2( \cot(2 \alpha ) - \tan(2 \alpha )) \\ = \tan( \alpha ) + 2 \cot(2 \alpha ) \\ = \tan( \alpha ) + \frac{2}{ \frac{2 \tan( \alpha ) }{1 - {( \tan( \alpha )) }^{2} } } \\ = \tan( \alpha ) + \frac{1 - { (\tan( \alpha ) )}^{2} }{ \tan( \alpha ) } \\ = \tan( \alpha ) + \cot( \alpha ) - \tan( \alpha ) \\ = \cot( \alpha )$