$ \cosec \theta - \sin \theta = p^3 $ and $ \sec \theta - \cos \theta = q^3 $
We are given two trigonometric equations:
We need to find the value of \(\tan \theta\).
Rewrite the equations using fundamental trigonometric identities (\(\ \cosec \theta = \frac{1}{\sin \theta}\), \(\ \sec \theta = \frac{1}{\cos \theta}\), \(\ 1 - \sin^2 \theta = \cos^2 \theta\), \(\ 1 - \cos^2 \theta = \sin^2 \theta\)).
To find \(\tan \theta\), we can divide Equation 2 by Equation 1:
\(\frac{q^3}{p^3} = \frac{(\frac{\sin^2 \theta}{\cos \theta})}{(\frac{\cos^2 \theta}{\sin \theta})}\)
Simplify the right side:
\(\frac{q^3}{p^3} = \frac{\sin^2 \theta}{\cos \theta} \times \frac{\sin \theta}{\cos^2 \theta}\)
\(\frac{q^3}{p^3} = \frac{\sin^3 \theta}{\cos^3 \theta}\)
Recognize that \(\ \frac{\sin^3 \theta}{\cos^3 \theta} = (\tan \theta)^3\):
\(\frac{q^3}{p^3} = (\tan \theta)^3\)
Take the cube root of both sides:
\(\tan \theta = \sqrt[3]{\frac{q^3}{p^3}}\)
\(\tan \theta = \frac{q}{p}\)
Therefore, \(\tan \theta\) is equal to \(\frac{q}{p}\). This matches Option B.
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