Solving for tan θ: Given Trigonometric Equations

We are given two trigonometric equations:

• \(\ \ \cosec \theta - \sin \theta = p^3\)
• \(\ \ \sec \theta - \cos \theta = q^3\)

We need to find the value of \(\tan \theta\).

Simplifying the Given Equations

Rewrite the equations using fundamental trigonometric identities (\(\ \cosec \theta = \frac{1}{\sin \theta}\), \(\ \sec \theta = \frac{1}{\cos \theta}\), \(\ 1 - \sin^2 \theta = \cos^2 \theta\), \(\ 1 - \cos^2 \theta = \sin^2 \theta\)).

1. \(\ \frac{1}{\sin \theta} - \sin \theta = p^3\)
   \(\implies \frac{1 - \sin^2 \theta}{\sin \theta} = p^3\)
   \(\implies \frac{\cos^2 \theta}{\sin \theta} = p^3\) (Equation 1)
2. \(\ \frac{1}{\cos \theta} - \cos \theta = q^3\)
   \(\implies \frac{1 - \cos^2 \theta}{\cos \theta} = q^3\)
   \(\implies \frac{\sin^2 \theta}{\cos \theta} = q^3\) (Equation 2)

Deriving tan θ

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:

\(\tan \theta = \sqrt[3]{\frac{q^3}{p^3}}\)

\(\tan \theta = \frac{q}{p}\)

Conclusion

Therefore, \(\tan \theta\) is equal to \(\frac{q}{p}\). This matches Option B.