If x = sec θ - cos θ and y = secⁿ θ - cosⁿ θ, then show that (x² + 4)(dy/dx)² = n²(y² + 4)

Q: Wait, why can't we just eliminate $\theta$ and write $y$ as a function of $x$?

You absolutely could, but it would be a trap! Solving for $\sec \theta$ in terms of $x$ involves solving a quadratic equation, and then raising that to the power of $n$ for $y$ creates a massive, unmanageable expression. The beauty of parametric differentiation is that it allows us to work with the derivatives directly, bypassing the need to explicitly define $y$ in terms of $x$.

Q: How did you know to factor out $\tan \theta$ in the derivatives?

That is a great observation! In trigonometric differentiation, especially with functions like $\sec \theta$ and $\cos \theta$, the derivatives often produce terms that share common factors. By factoring out $\tan \theta$, we reveal the underlying structure of the expression, which makes the subsequent division in the chain rule $\frac{dy/d\theta}{dx/d\theta}$ much cleaner. Always look for common factors when you see trigonometric derivatives!

Q: Is there a trick to remembering the identity $(A+B)^2 = (A-B)^2 + 4AB$?

It is not so much a trick as it is a fundamental algebraic relationship. Think of it this way: $(A-B)^2 = A^2 + B^2 - 2AB$. If you add $4AB$ to this, you get $A^2 + B^2 + 2AB$, which is exactly $(A+B)^2$. It is a lifesaver in problems involving reciprocals like $\sec \theta$ and $\cos \theta$ because their product is always $1$, making the $4AB$ term simply $4$.

Question

If $x = \sec 	heta - \cos 	heta$ and $y = \sec^n 	heta - \cos^n 	heta$, then show that $(x^2 + 4)\left(\frac{dy}{dx}ight)^2 = n^2(y^2 + 4)$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future IITians! Today, we are going to embark on a journey through a classic JEE Advanced problem that perfectly illustrates the power of parametric differentiation.

We are given two parametric equations: 
$$x = \sec 	heta - \cos 	heta$$ 
$$y = \sec^n 	heta - \cos^n 	heta$$

Our mission is to prove the beautiful relationship:
$$(x^2 + 4)\left(\frac{dy}{dx}ight)^2 = n^2(y^2 + 4)$$

This is not just a calculation; it is a testament to how trigonometry and algebra dance together in perfect harmony.

### The Parametric Chain Rule

Since $x$ and $y$ are both functions of the parameter $	heta$, we cannot directly find $\frac{dy}{dx}$. Instead, we must use the parametric chain rule, which states:
$$\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta}$$

Our first task is to find these two individual derivatives. Let us start with $x$. Differentiating $x = \sec 	heta - \cos 	heta$ with respect to $	heta$ gives us:
$$\frac{dx}{d	heta} = \sec 	heta 	an 	heta + \sin 	heta$$

Now, to make our lives easier, let us factor out $	an 	heta$. Since $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$, we can rewrite $\sin 	heta$ as $	an 	heta \cos 	heta$.

Thus, we obtain:
$$\frac{dx}{d	heta} = 	an 	heta (\sec 	heta + \cos 	heta)$$

### Differentiating $y$

Now, let us turn our attention to $y = \sec^n 	heta - \cos^n 	heta$. Applying the power rule and the chain rule, we get:
$$\frac{dy}{d	heta} = n \sec^{n-1} 	heta (\sec 	heta 	an 	heta) - n \cos^{n-1} 	heta (-\sin 	heta)$$

Simplifying this, we obtain:
$$\frac{dy}{d	heta} = n \sec^n 	heta 	an 	heta + n \cos^{n-1} 	heta \sin 	heta$$

Just as we did for $x$, let us factor out $n 	an 	heta$. This gives us:
$$\frac{dy}{d	heta} = n 	an 	heta (\sec^n 	heta + \cos^n 	heta)$$

Notice the symmetry! The structure of the derivative of $y$ mirrors the structure of the derivative of $x$.

### The Master Stroke of Algebra

Now, we substitute these into our parametric formula:
$$\frac{dy}{dx} = \frac{n 	an 	heta (\sec^n 	heta + \cos^n 	heta)}{	an 	heta (\sec 	heta + \cos 	heta)}$$

The $	an 	heta$ terms cancel out beautifully, leaving us with:
$$\frac{dy}{dx} = \frac{n (\sec^n 	heta + \cos^n 	heta)}{\sec 	heta + \cos 	heta}$$

Squaring both sides, we get:
$$\left(\frac{dy}{dx}ight)^2 = \frac{n^2 (\sec^n 	heta + \cos^n 	heta)^2}{(\sec 	heta + \cos 	heta)^2}$$

### Final Calculation

Now, we need to bridge the gap to our target equation. We use the identity $(A+B)^2 = (A-B)^2 + 4AB$.

For the denominator, let $A = \sec 	heta$ and $B = \cos 	heta$. Since $AB = 1$, we have:
$$(\sec 	heta + \cos 	heta)^2 = (\sec 	heta - \cos 	heta)^2 + 4 = x^2 + 4$$

Similarly, for the numerator, let $A = \sec^n 	heta$ and $B = \cos^n 	heta$. Again, $AB = 1$, so:
$$(\sec^n 	heta + \cos^n 	heta)^2 = y^2 + 4$$

Substituting these back, we get:
$$\left(\frac{dy}{dx}ight)^2 = \frac{n^2 (y^2 + 4)}{x^2 + 4}$$

Cross-multiplying gives us the final result:
$$(x^2 + 4)\left(\frac{dy}{dx}ight)^2 = n^2(y^2 + 4)$$

We have arrived at the proof! This problem is a brilliant reminder that when you see complex trigonometric expressions, look for the underlying algebraic structure.

Animated Solution for Mathematics - Differentiation: If $x = sec θ - cos θ$ and $y = sec^{n} θ - cos^{n} θ$ , then show that $(x^{2} + 4) (\frac{d y}{d x})^{2} = n^{2} (y^{2} + 4)$ .

Visualized Solution

Parametric Equations and Geometric Curve

Parametric Differentiation Rule

Differentiating $x$ w.r.t. $θ$

Factoring $\frac{d x}{d θ}$

Differentiating $y$ w.r.t. $θ$

Factoring $\frac{d y}{d θ}$

Finding and Squaring $\frac{d y}{d x}$

Algebraic Identity for $(x^{2} + 4)$

Algebraic Identity for $(y^{2} + 4)$

Final Substitution and Proof

The Sigma Insight: Techniques of Differentiation

Analyzing the Setup

The Parametric Chain Rule

Differentiating $y$

The Master Stroke of Algebra

Final Calculation

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Visualized Solution

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Parametric Differentiation Rule

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