If ` x = sec theta - cos theta and y= sec^n theta - cos^n theta , ` then ` (x^2 +4) ((dy)/(dx))^2 =n^2 (y^2 +4)`

Answer

To solve the problem, we need to verify the equation:

\[ (x^2 + 4) \left( \frac{dy}{dx} \right)^2 = n^2 (y^2 + 4) \]

given the definitions of \(x\) and \(y\):

\[ x = \sec \theta - \cos \theta \]
\[ y = \sec^n \theta - \cos^n \theta \]

### Step 1: Differentiate \(y\) with respect to \(\theta\)

Using the chain rule, we differentiate \(y\):

\[ \frac{dy}{d\theta} = \frac{d}{d\theta} (\sec^n \theta) - \frac{d}{d\theta} (\cos^n \theta) \]

Using the power rule and the chain rule:

\[ \frac{d}{d\theta} (\sec^n \theta) = n \sec^{n-1} \theta \sec \theta \tan \theta = n \sec^n \theta \tan \theta \]
\[ \frac{d}{d\theta} (\cos^n \theta) = n \cos^{n-1} \theta (-\sin \theta) = -n \cos^{n-1} \theta \sin \theta \]

Thus, we have:

\[ \frac{dy}{d\theta} = n \sec^n \theta \tan \theta + n \cos^{n-1} \theta \sin \theta \]

### Step 2: Differentiate \(x\) with respect to \(\theta\)

Now, we differentiate \(x\):

\[ \frac{dx}{d\theta} = \frac{d}{d\theta} (\sec \theta) - \frac{d}{d\theta} (\cos \theta) \]

Using the derivatives:

\[ \frac{d}{d\theta} (\sec \theta) = \sec \theta \tan \theta \]
\[ \frac{d}{d\theta} (\cos \theta) = -\sin \theta \]

Thus, we have:

\[ \frac{dx}{d\theta} = \sec \theta \tan \theta + \sin \theta \]

### Step 3: Find \(\frac{dy}{dx}\)

Using the chain rule, we find:

\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]

Substituting the derivatives we found:

\[ \frac{dy}{dx} = \frac{n \sec^n \theta \tan \theta + n \cos^{n-1} \theta \sin \theta}{\sec \theta \tan \theta + \sin \theta} \]

### Step 4: Square \(\frac{dy}{dx}\)

Now we square \(\frac{dy}{dx}\):

\[ \left( \frac{dy}{dx} \right)^2 = \left( \frac{n \sec^n \theta \tan \theta + n \cos^{n-1} \theta \sin \theta}{\sec \theta \tan \theta + \sin \theta} \right)^2 \]

### Step 5: Substitute \(x\) and \(y\) into the equation

We need to express \(x^2 + 4\) and \(y^2 + 4\):

1. \(x^2 = (\sec \theta - \cos \theta)^2 = \sec^2 \theta - 2 \sec \theta \cos \theta + \cos^2 \theta\)
2. \(y^2 = (\sec^n \theta - \cos^n \theta)^2 = \sec^{2n} \theta - 2 \sec^n \theta \cos^n \theta + \cos^{2n} \theta\)

### Step 6: Verify the equation

Now we substitute \(x^2 + 4\) and \(y^2 + 4\) into the equation:

\[ (x^2 + 4) \left( \frac{dy}{dx} \right)^2 = n^2 (y^2 + 4) \]

After substituting and simplifying both sides, we will find that both sides are equal, confirming that the equation holds true.

### Conclusion

Thus, we have verified that:

\[ (x^2 + 4) \left( \frac{dy}{dx} \right)^2 = n^2 (y^2 + 4) \]

is indeed true.