Question

please solve as soon as possible..​

Answer 1

Step-by-step explanation:

LHS

$\frac{ \sec(8 \alpha ) - 1}{ \sec(4 \alpha ) - 1 } \\ \\ = \frac{1 - \cos(8 \alpha ) }{1 - \cos(4 \alpha ) } \times \frac{ \cos(4 \alpha ) }{ \cos(8 \alpha ) } \\ \\ = \frac{2 { \sin(4 \alpha ) }^{2} }{2 { \sin(2 \alpha ) }^{2} } \times \frac{ \cos(4 \alpha ) }{ \cos(8 \alpha ) } \\ \\ \\ \\ = \frac{ 2\sin(4 \alpha ) \cos(4 \alpha ) \: \: \: \: ( \sin(4 \alpha ) )}{2 { \sin(2 \alpha ) }^{2} \cos(8 \alpha ) } \\ \\ = \frac{ \sin(8 \alpha ) }{ \cos(8 \alpha ) } \times \frac{2 \sin(2 \alpha ) \cos(2 \alpha ) }{2 { \sin(2 \alpha ) }^{2} } \\ \\ = \tan(8 \alpha ) \times \frac{ \cos(2 \alpha ) }{ \sin(2 \alpha ) } \\ \\ = \frac{ \tan(8 \alpha ) }{ \tan(2 \alpha ) }$

therefore LHS = RHS