Question

If cosecθ - sinθ = m and secθ - cosθ = n, then prove (m²n)2/3 + (mn²)2/3 = 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 ]

Answer 2

hope this will help you