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


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