If ` x =asec theta,y = atan theta , then" at" theta =(pi)/( 4) ,(dy)/(dx) =`

Options

A. ` sqrt2`
B. `2`
C. ` (1)/(sqrt(2))`
D. ` (1)/(2) `

Answer

To solve the problem, we need to find \(\frac{dy}{dx}\) when \(x = a \sec \theta\) and \(y = a \tan \theta\) at \(\theta = \frac{\pi}{4}\).

### Step-by-Step Solution:

1. **Differentiate \(x\) with respect to \(\theta\)**:
\[ x = a \sec \theta \]
The derivative of \(\sec \theta\) is \(\sec \theta \tan \theta\). Therefore,
\[ \frac{dx}{d\theta} = a \sec \theta \tan \theta \]

2. **Differentiate \(y\) with respect to \(\theta\)**:
\[ y = a \tan \theta \]
The derivative of \(\tan \theta\) is \(\sec^2 \theta\). Therefore,
\[ \frac{dy}{d\theta} = a \sec^2 \theta \]

3. **Find \(\frac{dy}{dx}\)**:
We use the chain rule to find \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{a \sec^2 \theta}{a \sec \theta \tan \theta} \]
The \(a\) terms cancel out:
\[ \frac{dy}{dx} = \frac{\sec^2 \theta}{\sec \theta \tan \theta} \]
Simplifying further:
\[ \frac{dy}{dx} = \frac{\sec \theta}{\tan \theta} \]

4. **Express \(\frac{dy}{dx}\) in terms of sine and cosine**:
Recall that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and \(\sec \theta = \frac{1}{\cos \theta}\):
\[ \frac{dy}{dx} = \frac{\sec \theta}{\tan \theta} = \frac{1/\cos \theta}{\sin \theta/\cos \theta} = \frac{1}{\sin \theta} = \csc \theta \]

5. **Evaluate \(\frac{dy}{dx}\) at \(\theta = \frac{\pi}{4}\)**:
We know that \(\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\):
\[ \frac{dy}{dx} \bigg|_{\theta = \frac{\pi}{4}} = \csc\left(\frac{\pi}{4}\right) = \frac{1}{\sin\left(\frac{\pi}{4}\right)} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2} \]

### Final Answer:
\[ \frac{dy}{dx} \bigg|_{\theta = \frac{\pi}{4}} = \sqrt{2} \]