Good question

Question

cosec theta-sin theta =m

sec theta-cos theta =n

Show: (m^2 n) ^2/3 + (mn^2)^2/3=1

 

 

Answer 1

m = cscθ - sinθ = 1/sinθ - sinθ = (1-sin²θ)/sinθ = cos²θ/sinθ

 

Similarly, n = sin²θ / cosθ

 

So, (m²n)^2/3 + (mn²)^2/3

 

   = [(cos⁴θ/sin²θ) (sin²θ/cosθ)]^2/3 + [(cos²θ/sinθ)(sin⁴θ/cos²θ]^2/3

 

   = (cos³θ)^2/3 + (sin³θ)^2/3 = cos²θ + sin²θ = 1