Question

if cosecθ - sinθ = m and secθ - tanθ = n , prove that (m2n)2/3 + (mn2)2/3 = 1

Answer 1

here , you did mistake in
sec∅ - tan∅ = n , here should be sec∅ - cos∅ = n

I mean question should be ,
if cosec∅ - sin∅ = m
sec∅ - cos∅ = n
then, proved that (m²n)^(2/3) + (mn²)^(2/3) = 1



cosec∅ - sin∅ = m
1/sin∅ - sin∅ = m
(1 - sin²∅)/sin∅ = m
(cos²∅)/sin∅ = m ---------------(1)

again,
sec∅ - cos∅ = n
1/cos∅ - cos∅ = n
(1 - cos²∅)/cos∅ = n
sin²∅/cos∅ = n ------------(2)

now,
m²n = {cos²∅/sin∅}² × {sin²∅/cos∅}
= cos⁴∅/sin²∅ × sin²∅/cos∅
= cos³∅

so, (m²n)^(2/3) = (cos³∅)^(2/3) = cos²∅ -----(3)

again,
mn² = {cos²∅/sin∅} × {sin²∅/cos∅}
= cos²∅/sin∅ × sin⁴∅/cos²∅
= sin³∅

so , (mn²)^(2/3)=(sin³∅)^2/3 =sin²∅ -----(4)

from equations (3) and (4)
we get,
(m²n)^(2/3) + (mn²)^(2/3) = cos²∅ + sin²∅ = 1

hence ,
(m²n)^(2/3) + (mn²)^(2/3) = 1

Answer 2

use this
I think ur question was little incorrect