If `x=sect+tant and y=sect - tant`, where t is a parameter, then the value of `(dy)/(dx)` when `x=(1)/(sqrt3)` is

Options

A. 0
B. `-3`
C. `sqrt3`
D. `(1)/(sqrt3)`

Answer

To solve the problem, we need to find the value of \(\frac{dy}{dx}\) given the equations \(x = \sec t + \tan t\) and \(y = \sec t - \tan t\) when \(x = \frac{1}{\sqrt{3}}\).

### Step-by-Step Solution:

1. **Set up the equations**:
We have:
\[ x = \sec t + \tan t \]
\[ y = \sec t - \tan t \]

2. **Substitute the value of \(x\)**:
We substitute \(x = \frac{1}{\sqrt{3}}\):
\[ \frac{1}{\sqrt{3}} = \sec t + \tan t \]

3. **Use the identity**:
We know that:
\[ \sec^2 t - \tan^2 t = 1 \]
This can be factored as:
\[ (\sec t - \tan t)(\sec t + \tan t) = 1 \]
Substituting \(x\) into the equation, we get:
\[ y \cdot \frac{1}{\sqrt{3}} = 1 \]

4. **Solve for \(y\)**:
Rearranging gives:
\[ y = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3} \]

5. **Differentiate \(y\) and \(x\)**:
We need to find \(\frac{dy}{dx}\). Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

6. **Differentiate \(y\)**:
Differentiate \(y = \sec t - \tan t\):
\[ \frac{dy}{dt} = \sec t \tan t - \sec^2 t \]

7. **Differentiate \(x\)**:
Differentiate \(x = \sec t + \tan t\):
\[ \frac{dx}{dt} = \sec t \tan t + \sec^2 t \]

8. **Substitute into \(\frac{dy}{dx}\)**:
Now substituting into \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{\sec t \tan t - \sec^2 t}{\sec t \tan t + \sec^2 t} \]

9. **Substitute known values**:
We know:
- \(y = \sqrt{3}\) (from step 4)
- \(x = \frac{1}{\sqrt{3}}\) (from step 2)

We can express \(\sec t + \tan t = \frac{1}{\sqrt{3}}\) and \(\sec t - \tan t = \sqrt{3}\).

10. **Find \(\sec t\) and \(\tan t\)**:
Let:
\[ \sec t + \tan t = \frac{1}{\sqrt{3}} \quad (1) \]
\[ \sec t - \tan t = \sqrt{3} \quad (2) \]
Adding (1) and (2):
\[ 2\sec t = \frac{1}{\sqrt{3}} + \sqrt{3} \]
\[ \sec t = \frac{1 + 3}{2\sqrt{3}} = \frac{2}{\sqrt{3}} \]
Substituting \(\sec t\) back to find \(\tan t\):
\[ \tan t = \sec t - \sqrt{3} = \frac{2}{\sqrt{3}} - \sqrt{3} \]

11. **Final calculation**:
Substitute \(\sec t\) and \(\tan t\) into \(\frac{dy}{dx}\) and simplify:
\[ \frac{dy}{dx} = \frac{\left(\frac{2}{\sqrt{3}} \cdot \tan t - \left(\frac{2}{\sqrt{3}}\right)^2\right)}{\left(\frac{2}{\sqrt{3}} \cdot \tan t + \left(\frac{2}{\sqrt{3}}\right)^2\right)} \]
After simplification, we find that:
\[ \frac{dy}{dx} = -3 \]

Thus, the value of \(\frac{dy}{dx}\) when \(x = \frac{1}{\sqrt{3}}\) is \(-3\).