Prove that.... Plz answer of this question.
Attachments:
Answers
Answered by
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
Attachments:
Similar questions