Sin⁴A+sin²A=1 then prove that 1/tan⁴A+1/tan²A=1
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sin⁴A+sin²A=1
or, sin⁴A=1-sin²A
or, sin⁴A=cos²A ------------------------(1)
∴, 1/tan⁴A+1/tan²A
=1/(sin⁴A/cos⁴A)+1/(sin²A/cos²A)
=cos⁴A/sin⁴A+cos²A/sin²A
=(cos⁴A+sin²Acos²A)/sin⁴A
=cos²A(cos²A+sin²A)/sin⁴A
=cos²A/sin⁴A [∵, sin²A+cos²A=1]
=cos²A/cos²A [using (1)]
=1 (Proved)
or, sin⁴A=1-sin²A
or, sin⁴A=cos²A ------------------------(1)
∴, 1/tan⁴A+1/tan²A
=1/(sin⁴A/cos⁴A)+1/(sin²A/cos²A)
=cos⁴A/sin⁴A+cos²A/sin²A
=(cos⁴A+sin²Acos²A)/sin⁴A
=cos²A(cos²A+sin²A)/sin⁴A
=cos²A/sin⁴A [∵, sin²A+cos²A=1]
=cos²A/cos²A [using (1)]
=1 (Proved)
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