Forum_Post_Question_2.JPG
By rewriting the equations:
\(\frac {1}{sin \theta} - sin \theta = m\) - equation 1
\(\frac {1}{cos \theta} - cos \theta = n\) - equation 2
\(\frac {1 - sin^2 \theta}{sin \theta}\)=\(\frac {cos^2 \theta}{sin \theta}\) =m
\(\frac {1- cos^2 \theta}{cos \theta}\)=\(\frac {sin^2 \theta}{cos \theta}\) =n
(m2n)2/3=\(\frac {cos^4 \theta}{sin^2 \theta}\) x \(\frac {sin^2 \theta}{cos \theta}
= cos^2 \theta\)
(mn2)2/3=\(\frac {sin^4 \theta}{cos^2 \theta}\) x \(\frac {cos^2 \theta}{sin \theta}
= sin^2 \theta\)
(m2n)2/3 + (mn2)2/3= cos2 \(\theta\) + sin2 \(\theta\) = 1
\(\frac {1}{sin \theta} - sin \theta = m\) - equation 1
\(\frac {1}{cos \theta} - cos \theta = n\) - equation 2
\(\frac {1 - sin^2 \theta}{sin \theta}\)=\(\frac {cos^2 \theta}{sin \theta}\) =m
\(\frac {1- cos^2 \theta}{cos \theta}\)=\(\frac {sin^2 \theta}{cos \theta}\) =n
(m2n)2/3=\(\frac {cos^4 \theta}{sin^2 \theta}\) x \(\frac {sin^2 \theta}{cos \theta}
= cos^2 \theta\)
(mn2)2/3=\(\frac {sin^4 \theta}{cos^2 \theta}\) x \(\frac {cos^2 \theta}{sin \theta}
= sin^2 \theta\)
(m2n)2/3 + (mn2)2/3= cos2 \(\theta\) + sin2 \(\theta\) = 1
Thank you very much!