If `x = a^(2)(sin theta + "cosec" theta) y = a^(2) (sin theta - "cosec" theta)`, then `(dy)/(dx)` =

Options

A. `y/x`
B. `(-y)/x`
C. `x/y`
D. `(-x)/y`

Answer

To find \(\frac{dy}{dx}\) given the equations \(x = a^2(\sin \theta + \csc \theta)\) and \(y = a^2(\sin \theta - \csc \theta)\), we will follow these steps:

### Step 1: Differentiate \(x\) with respect to \(\theta\)
We start with the expression for \(x\):
\[ x = a^2(\sin \theta + \csc \theta) \]
To differentiate \(x\) with respect to \(\theta\), we apply the product rule and the chain rule. The derivative of \(\sin \theta\) is \(\cos \theta\) and the derivative of \(\csc \theta\) is \(-\csc \theta \cot \theta\):
\[ \frac{dx}{d\theta} = a^2\left(\cos \theta - \csc \theta \cot \theta\right) \]

### Step 2: Differentiate \(y\) with respect to \(\theta\)
Next, we differentiate \(y\):
\[ y = a^2(\sin \theta - \csc \theta) \]
Using the same differentiation rules:
\[ \frac{dy}{d\theta} = a^2\left(\cos \theta + \csc \theta \cot \theta\right) \]

### Step 3: Find \(\frac{dy}{dx}\)
Now, we can find \(\frac{dy}{dx}\) using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{a^2\left(\cos \theta + \csc \theta \cot \theta\right)}{a^2\left(\cos \theta - \csc \theta \cot \theta\right)} \]
The \(a^2\) terms cancel out:
\[ \frac{dy}{dx} = \frac{\cos \theta + \csc \theta \cot \theta}{\cos \theta - \csc \theta \cot \theta} \]

### Step 4: Simplify the expression
We can rewrite \(\csc \theta\) as \(\frac{1}{\sin \theta}\) and \(\cot \theta\) as \(\frac{\cos \theta}{\sin \theta}\):
\[ \frac{dy}{dx} = \frac{\cos \theta + \frac{1}{\sin \theta} \cdot \frac{\cos \theta}{\sin \theta}}{\cos \theta - \frac{1}{\sin \theta} \cdot \frac{\cos \theta}{\sin \theta}} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{\cos \theta + \frac{\cos \theta}{\sin^2 \theta}}{\cos \theta - \frac{\cos \theta}{\sin^2 \theta}} \]

### Step 5: Factor out \(\cos \theta\)
Factoring \(\cos \theta\) from both the numerator and denominator:
\[ \frac{dy}{dx} = \frac{\cos \theta\left(1 + \frac{1}{\sin^2 \theta}\right)}{\cos \theta\left(1 - \frac{1}{\sin^2 \theta}\right)} \]
Canceling \(\cos \theta\):
\[ \frac{dy}{dx} = \frac{1 + \frac{1}{\sin^2 \theta}}{1 - \frac{1}{\sin^2 \theta}} \]

### Final Result
Thus, the final result for \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{\sin^2 \theta + 1}{\sin^2 \theta - 1} \]