Question

If cosecθ - sinθ = m and secθ - cosθ = n, then prove (m²n)2/3 + (mn²)2/3 = 1

Answer 1

cosecθ - sinθ = m
can be written as
1/ sinθ - sinθ = m
(1- (sinθ)^2 )/ sinθ = m
or
(cosθ)^2 / sinθ = m
similarly
(sinθ)^2 )/ cosθ=n
and so on...

Answer 2

I'm having confusion with the raised to 2/3 part.

Answer 3

ok

Answer 4

now find
m^2* n = ((cosθ)^2 / sinθ )^2 * (sinθ)^2 )/ cosθ = (cosθ)^3

Answer 5

similarly
n^2*m = ((sinθ)^2 )/ cosθ )^2 *(cosθ)^2 / sinθ = (sinθ)^3
can you do now

Answer 6

(m^2* n )^(2/3) = ((cosθ)^3 )^(2/3)=(cosθ)^2 and
(n^2* m )^(2/3) = ((sinθ)^3 )^(2/3)=(sinθ)^2
so adding we have
(m^2* n )^(2/3) + (n^2* m )^(2/3) = (cosθ)^2 + (sinθ)^2
=1

Answer 7

if that helps clear any confusion ^^