Math, asked by Anonymous, 10 months ago

Prove that.... Plz answer of this question. ​

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Answered by ShuchiRecites
23

Solution: (11) sec⁴A = 1/cos⁴A

→ sec⁴A(1 - sin⁴A) - 2tan²A = L.H.S

→ (1 - sin⁴A)/cos⁴A - 2 sin²A/cos²A

→ [(1 - sin⁴A) - 2sin²Acos²A]/cos⁴A

→ (1 - sin⁴A - 2sin²Acos²A)/cos⁴A

→ [1 - (sin²A)² - 2 sin²A cos²A]/cos⁴A

→ [1 - (1 - cos²A)² - 2sin²A cos²A]/cos⁴A

→ [1 - (1 + cos⁴A - 2cos²A) - 2sin²A cos²A]/cos⁴A

→ (2cos²A - 2sin²A cos²A - cos⁴A)/cos⁴A

→ [2cos²A(1 - sin²A) - cos⁴A]/cos⁴A

→ [2cos²A × cos²A - cos⁴A]/cos⁴A

→ [2cos⁴A - cos⁴A]/cos⁴A

→ cos⁴A/cos⁴A = 1 = R.H.S

(12) Refer to attachment

Q.E.D

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