If (cot θ + tan θ) = m and (sec θ – cos θ) = n, prove that (m2n) 2/3 – (mn2) 2/3 = 1.
Given: (cot θ + tan θ) = m
(sec θ – cos θ) = n
Since, m = cot θ + tan θ
= (1/tan θ) + tan θ
=
= sec2 θ/tan θ
= 1/(sin θ cos θ)
Also, n = sec θ – cos θ
= (1/cos θ) – cos θ
= (1 – cos2 θ)/cos θ
= sin2 θ/cos θ
Now, consider the left – hand side:
(m2n) 2/3 – (mn2) 2/3 =
=
=
=
= (1 – sin2 θ)cos2 θ
= cos2 θ/ cos2 θ
= 1