Question

If cosecA- sinA =m and secA- cosA= n. Prove that (m^2n)^2/3 + (mn^2)^2/3= 1.

Answer 1

We have
cosecA - SinA = m and
SecA - CosA = n
1/sinA - Sin = m and
1/cosA - cosA- = n
1 - sin²A/SinA = m and
1 - Cos²/CosA = n
cos² A/sinA = m and
Sin²A/CosA = n
(since; (1 - sin²A = cos² A and
1 - Cos² = Sin²A )

(mn²)⅔ + (m 2n)⅔ = (cos²A/sinA× sin⁴A/cos²A)⅔ + (cos4A/sin2A× sin2A/cosA)⅔

= (sin³A)⅔ + (cos³A)⅔

= (sin²A+cos²A) =1

Therefore (mn²)⅔ + (m²n)⅔ =1 (proved)
Hence L.H.S=R.H.S