Question

sin 3A/sinA – cos3A/cosA​

Answer 1

Answer:

$\frac{ \sin(3 \alpha ) }{ \sin( \alpha ) } - \frac{ \cos(3 \alpha ) }{ \cos( \alpha ) } \\ = \frac{3 \sin( \alpha ) - 4 { \sin( \alpha ) }^{3} }{ \sin( \alpha ) } - \frac{4 \ { \cos( \alpha ) }^{3} - 3 \cos( \alpha ) }{ \cos( \alpha ) } \\ = now \: take \: common \: | \sin( \alpha ) | and \: | \cos( \alpha ) | then \: cancell \: it \: with \: down \: one \\ = 3 - 4 { \sin( \alpha ) }^{2} - 4 { \cos( \alpha ) }^{2} + 3 \\ = 6 - 4( { \ \sin ( \alpha ) }^{2} + { \cos( \alpha ) }^{2} \\ = 6 - 4 \times 1 \\ = 6 - 4 \\ = 2$

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