If \(\frac{cos^2 \ \theta}{sin \ \theta} = p\) and \(\frac{sin^2 \ \theta}{cos \ \theta} = q\), then prove that \(p^2 q^2 (p^2 + q^2 + 3) = 1\).
Proof:
\(\frac{cos^2 \ \theta}{sin \ \theta} = p\) and \(\frac{sin^2 \ \theta}{cos \ \theta} = q\)
\(p^2=\)
\(q^2=\)
\(p^2 q^2=\)
\(p^2 q^2 (p^2 + q^2 + 3) =\)
By simplyfing this then applying the formula then we get the result.
Answer variants:
\(a^3+b^3=(a+b)^3-3ab(a^2+ab+b^2\)
\(cos^6 \theta+sin^6 \theta+3sin^2 \theta cos^2 \theta}\)
\(\frac{sin^4 \theta}{cos^2 \theta}\)
\(\frac{cos^4 \theta}{sin^2 \theta}\)
\(cos^6 \theta+sin^6 \theta-3sin^2 \theta cos^2 \theta}\)
\(a^3-b^3=(a-b)^3+3ab(a^2+ab+b^2\)
\(cos^2 \theta sin^2 \theta}\)
\(\frac{sin^2 \theta}{cos^2 \theta}\)
\(\frac{cos^2 \theta}{sin^2 \theta}\)