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Question

$$\frac{\ secA -\ tanA}{ \ secA + \ tanA} = 1 - 2\ secA \ tanA + 2\ tan^2A$$


Solution

$$LHS=\cfrac { \sec { A } -\tan { A }  }{ \sec { A } +\tan { A }  } \\ =\cfrac { \sec { A } -\tan { A }  }{ \sec { A } +\tan { A }  } \times \cfrac { \sec { A } -\tan { A }  }{ \sec { A } -\tan { A }  } \\ =\cfrac { \sec { ^{ 2 }A } -2\sec { A } \tan { A } +\tan { ^{ 2 }A }  }{ \sec { ^{ 2 }A } -\tan { ^{ 2 }A }  } \\ =1+\tan { ^{ 2 }A } -2\sec { A } \tan { A } +\tan { ^{ 2 }A } \\ =1-2\sec { A } \tan { A } +2\tan { ^{ 2 }A } $$

Mathematics

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