Question

if cosecθ-sinθ =m^3 and secθ-cosθ=n^3 then prove that n^2 m^4 +n^4 m^2 = 1

Answer 1

cosec∅ - sin∅ = m -----(1)

sec∅ - cos∅ = n --------(2)

from equation ,(1)

cosec∅ - sin∅ = m

1/sin∅ - sin∅ = m

1 - sin²∅ = m.sin∅

sin²∅ + m.sin∅ -1 = 0

m.sin∅ = 1 - sin²∅ = cos²∅

m = cos²∅/sin∅ ---------(3)

again, similarly , from equation ,(2)

sec∅ - cos∅ = n

1/cos∅ - cos∅ = n

1 - cos²∅ = n.cos∅

cos²∅ + n.cos∅ -1 = 0

n.cos∅ = 1 - cos²∅ = sin²∅

n = sin²∅/cos∅ -------(4)

put equation (3) and (4) in

(m²n)⅔ + (mn²)⅔

LHS = {(cos²∅/sin∅)²(sin²∅/cos∅)}⅔ + {(cos²∅/sin∅)(sin²∅/cos∅)²}⅔

= { cos⁴∅ × sin²∅/sin²∅ × cos∅}⅔ + { cos²∅×sin⁴∅/sin∅×cos²∅}⅔

= { cos³∅}⅔ + {sin³∅}⅔

= cos²∅ + sin²∅ = 1 = RHS

[ we know , sin²x + cos²x = 1 ]

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