if sin^4A+sin^2 A=1, prove that 1/tan^4A+1/tan^2 A=1
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Given that
sin⁴A+sin²A=1
=> sin⁴A=1-sin²A
=> sin⁴A=cos²A [1-sin²A=cos²A]
=> sin⁴A=cos²A .....................(equation i )
=> sin²A*sin²A=cos²A
=> sin²A=cos²A/sin²A
=> sin²A=cot²A [cos²A/sin²A=cot²A]
=> sin²A=1/tan²A [ cotA=1/tanA]
=> sin²A=1/tan²A .......................( equation ii )
Now,
Multiplying cos2A on both sides in equation i,
=> sin⁴A*cos²A=cos²A*cos²A
=> sin⁴Acos²A=cos⁴A
=> cos²A=cos⁴A/sin⁴A
=> cos²A=cot⁴A [ Since, cos²A/sin²A=cot²A]
=> cos²A=1/tan⁴A [ Since, cot²A=1/tan²A ]
=> cos²A=1/tan⁴A ....................( equation iii )
Adding equation (ii) and (iii),
sin²A=1/tan²A
(+) cos²A = 1/tan⁴A
--------------------------------------------------
cos²A+sin²A=1/tan²A+1/tan⁴A
=> 1/tan²A+1/tan⁴A=cos²A+sin²A
=> 1/tan²A+1/tan⁴A=1 [ Since, cos²A+sin²A=1 ]
=> 1/tan²A+1/tan⁴A=1 [Proved]
sin⁴A+sin²A=1
=> sin⁴A=1-sin²A
=> sin⁴A=cos²A [1-sin²A=cos²A]
=> sin⁴A=cos²A .....................(equation i )
=> sin²A*sin²A=cos²A
=> sin²A=cos²A/sin²A
=> sin²A=cot²A [cos²A/sin²A=cot²A]
=> sin²A=1/tan²A [ cotA=1/tanA]
=> sin²A=1/tan²A .......................( equation ii )
Now,
Multiplying cos2A on both sides in equation i,
=> sin⁴A*cos²A=cos²A*cos²A
=> sin⁴Acos²A=cos⁴A
=> cos²A=cos⁴A/sin⁴A
=> cos²A=cot⁴A [ Since, cos²A/sin²A=cot²A]
=> cos²A=1/tan⁴A [ Since, cot²A=1/tan²A ]
=> cos²A=1/tan⁴A ....................( equation iii )
Adding equation (ii) and (iii),
sin²A=1/tan²A
(+) cos²A = 1/tan⁴A
--------------------------------------------------
cos²A+sin²A=1/tan²A+1/tan⁴A
=> 1/tan²A+1/tan⁴A=cos²A+sin²A
=> 1/tan²A+1/tan⁴A=1 [ Since, cos²A+sin²A=1 ]
=> 1/tan²A+1/tan⁴A=1 [Proved]
vanshgupta108vp85967:
very confusing
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