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guapka [62]
2 years ago
9

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

Mathematics
1 answer:
shepuryov [24]2 years ago
8 0
LHS\\ \\ =\frac { \tan ^{ 2 }{ x-\sin ^{ 2 }{ x }  }  }{ \tan { x }  } \\ \\ =\frac { 1 }{ \tan { x }  } \left( \tan ^{ 2 }{ x-\sin ^{ 2 }{ x }  }  \right)

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

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

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