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MATHEMATICS HIGH SCHOOL

How do I get (tan^2(x)-sin^2(x))/tan(x) equal to (sin^2(x))/cot(x)

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Answer 1
Answer: LHS\n \n =\frac { \tan ^( 2 ){ x-\sin ^( 2 ){ x } } }{ \tan { x } } \n \n =\frac { 1 }{ \tan { x } } \left( \tan ^( 2 ){ x-\sin ^( 2 ){ x } } \right)

\n \n =\frac { \cos { x } }{ \sin { x } } \left( \frac { \sin ^( 2 ){ x } }{ \cos ^( 2 ){ x } } -\frac { \sin ^( 2 ){ x\cos ^( 2 ){ x } } }{ \cos ^( 2 ){ x } } \right) \n \n =\frac { \cos { x } }{ \sin { x } } \left( \frac { \sin ^( 2 ){ x-\sin ^( 2 ){ x\cos ^( 2 ){ x } } } }{ \cos ^( 2 ){ x } } \right)

\n \n =\frac { \cos { x } }{ \sin { x } } \cdot \frac { \sin ^( 2 ){ x\left( 1-\cos ^( 2 ){ x } \right) } }{ \cos ^( 2 ){ x } } \n \n =\frac { \cos { x } }{ \sin { x } } \cdot \frac { \sin ^( 2 ){ x\cdot \sin ^( 2 ){ x } } }{ \cos ^( 2 ){ x } } \n \n =\frac { \cos { x } \sin ^( 4 ){ x } }{ \sin { x\cos ^( 2 ){ x } } } \n \n =\frac { \sin ^( 3 ){ x } }{ \cos { x } }

\n \n =\sin ^( 2 ){ x } \cdot \frac { \sin { x } }{ \cos { x } } \n \n =\sin ^( 2 ){ x } \cdot \frac { 1 }{ \frac { \cos { x } }{ \sin { x } } } \n \n =\sin ^( 2 ){ x } \cdot \frac { 1 }{ \cot { x } } \n \n =\frac { \sin ^( 2 ){ x } }{ \cot { x } } \n \n =RHS

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