Question

Prove tan³Q/1+tan²Q+cot³Q/1+cot²Q= secQcosecQ-2sinQcosQ

Answer 1

$\frac{ { \tan( \alpha ) }^{3} }{1 + { \tan( \alpha ) }^{2} } + \frac{ { \cot( \alpha ) }^{3} }{1 + { \cot( \alpha ) }^{2} } = \sec( \alpha ) \csc( \alpha ) - 2 \sin( \alpha ) \cos( \alpha ) \\ lhs \\ \frac{ { \tan( \alpha ) }^{3} }{1 + { \tan( \alpha ) }^{2} } + \frac{ { \cot( \alpha ) }^{3} }{1 + { \cot( \alpha ) }^{2} } \\ = \frac{ \frac{ { \sin( \alpha ) }^{3} }{ { \cos( \alpha ) }^{3} } }{ { \sec( \alpha ) }^{2} } + \frac{ \frac{ { \cos( \alpha ) }^{3} }{ { \sin( \alpha ) }^{3} } }{ { \csc( \alpha ) }^{2} } \\ = \frac{ { \sin( \alpha ) }^{3} }{ \cos( \alpha ) } + \frac{ { \cos( \alpha ) }^{3} }{ \sin( \alpha ) } \\ = \frac{ { \sin( \alpha) }^{4} + { \cos( \alpha ) }^{4} }{ \sin( \alpha ) \cos( \alpha ) } \\ = {( { \sin( \alpha ) }^{2} + { \cos( \alpha ) }^{2} ) }^{2} - 2 { \sin( \alpha ) }^{2} { \cos( \alpha ) }^{2} \div \sin( \alpha ) \cos( \alpha ) \\ = \frac{1 - 2 { \sin( \alpha ) }^{2} { \cos( \alpha ) }^{2} }{ \sin( \alpha ) \cos( \alpha ) } \\ = \frac{1}{ \sin( \alpha )\cos( \alpha ) } - \frac{2 { \sin( \alpha ) }^{2} { \cos( \alpha ) }^{2} }{ \sin( \alpha ) \cos( \alpha ) } \\ = \sec( \alpha ) \csc( \alpha ) - 2 \sin( \alpha ) \cos( \alpha ) \\ = rhs$

Answer 2

Answer:

Refer to the attachement...