Q6 of 45 Page 131

Prove that:

sec4A (1– sin4A) – 2tan2 A = 1

Taking LHS


= sec4A(1 - sin4A) - 2tan2A


= sec4A - sin4A sec4A - 2tan2A



= sec4A - tan4A - tan2A - tan2A


= sec4A - tan2A(1 + tan2A) - tan2A


= sec4A - tan2A sec2A - tan2A [As, sec2θ = 1 + tan2θ ]


= sec2A(sec2A - tan2A) - tan2A


= sec2A - tan2A [As, sec2θ = 1 + tan2θ ⇒ sec2θ - tan2θ = 1]


= 1


= RHS


Proved !


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