Question

answer this question plz

Answer 1

Heya

$\frac{ \sin( \alpha ) - { 2\sin( \alpha ) }^{3} }{2 { \cos( \alpha ) }^{3} - \cos( \alpha ) } = \frac{ \sin( \alpha ) }{ \cos( \alpha ) } \times \frac{1 - { 2\sin( \alpha ) }^{2} }{2 { \cos( \alpha ) }^{2} - 1 }$


Now,

$1 - { 2\sin( \alpha ) }^{2} = \cos( \alpha^{2} ) - { \sin( \alpha ) }^{2}$

Similarly, the denominator also breaks to :
$2 { \cos( \alpha ) }^{2} - 1 = { \cos( \alpha ) }^{2} - { \sin( \alpha ) }^{2}$

Hence, our expression reduces to :
$\frac{ \sin( \alpha ) }{ \cos( \alpha ) } \times \frac{1 - { 2\sin( \alpha ) }^{2} }{2 { \cos( \alpha ) }^{2} - 1 } = \frac{ \sin( \alpha ) }{ \cos( \alpha ) } \times \frac{ { \cos( \alpha ) }^{2} - { \sin( \alpha ) }^{2} }{ { \cos( \alpha ) }^{2} - { \sin( \alpha ) }^{2} }$

$= \frac{ \sin( \alpha ) }{ \cos( \alpha ) } = \tan( \alpha )$