If cot A + tan A = x ,sec A - cos A = y Then prove that (x 2 y) 2/3 _ (xy 2 ) 2/3 = 1
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x= cosa/sina + sina/cosa
=(cos2a + sin2a) / sinacosa
= 1/(sinacosa)
y= 1/cosa - cosa
=1-cos2a/cosa
=sin2a/cosa
now, LHS = (x2y)2/3 - (xy2)2/3
= [(1/sinacosa)2* sin2a/cosa]2/3 - [(1/sinacosa * (sin2a/cosa)2)]2/3
= [1/sin2acos2a * sin2a / cosa]2/3 - [1/sinacosa * sin4a/ cos2a]2/3
=[1/ cos3a] 2/3 - [sin3a/co3a]2/3 (.`.after cancelling out common terms)
= (sec3a)2/3 - (tan3a)2/3 (`.` (am)n = amn)
= sec2a - tan2a
=1
= RHS
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