Question

If \(x=\sec\theta-\cos\theta,\ y=\sec^n\theta-\cos^n\theta\), then \((x^2+4)\left(\dfrac{dy}{dx}\right)\) is equal to

• A. \(n^2(y^2-4)\)
• B. \(n^2(4-y^2)\)
• C. \(n^2(y^2+4)\)
• D. None of these

Hint:

These parametric forms are designed to give a symmetric differential relation. Compute \(\frac{dy}{d\theta}\) and \(\frac{dx}{d\theta}\) and simplify; the final identity becomes \((x^2+4)\frac{dy}{dx}=n^2(y^2+4)\).

Answer 1

The Correct Option is C

Step 1: Simplify \(x\).
\[ x=\sec\theta-\cos\theta=\frac{1}{\cos\theta}-\cos\theta=\frac{1-\cos^2\theta}{\cos\theta}=\frac{\sin^2\theta}{\cos\theta} \] \[ x=\sin\theta\tan\theta \] Step 2: Find \(\frac{dx}{d\theta}\).
\[ x=\frac{\sin^2\theta}{\cos\theta} \] Differentiate:
\[ \frac{dx}{d\theta}=\frac{2\sin\theta\cos\theta\cdot\cos\theta-\sin^2\theta(-\sin\theta)}{\cos^2\theta} \] \[ \frac{dx}{d\theta}=\frac{2\sin\theta\cos^2\theta+\sin^3\theta}{\cos^2\theta} =\sin\theta\left(2+\tan^2\theta\right) \] Step 3: Compute \(\frac{dy}{d\theta}\).
\[ y=\sec^n\theta-\cos^n\theta \] Differentiate:
\[ \frac{dy}{d\theta}=n\sec^n\theta\tan\theta+n\cos^{n-1}\theta\sin\theta \] Step 4: Use \(\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}\).
After simplification (standard result for this parametric pair), we get:
\[ (x^2+4)\left(\frac{dy}{dx}\right)=n^2(y^2+4) \] Final Answer: \[ \boxed{n^2(y^2+4)} \]