If `x=int_(0)^(t^(2))e^(sqrt(z)){(2tan sqrt(z)+1-tan^(2)sqrt(z))/(2sqrt(z)sec^(2)sqrt(z))}dz` and `x=int_(0)^(t^(2))e^(sqrt(z)){(1-tan^(2)sqrt(z)-2tan sqrt(z))/(2sqrt(z)sec^(2)sqrt(z))}dz` : Then the inclination of the tangent to the curve at `t=(pi)/(4)` is :

Options

A. `(pi)/(4)`
B. `(pi)/(3)`
C. `(pi)/(2)`
D. `(3pi)/(4)`

Answer

To solve the given problem, we need to find the inclination of the tangent to the curve at \( t = \frac{\pi}{4} \). We are given two integrals for \( x \) and \( y \):

1. \( x = \int_{0}^{t^2} e^{\sqrt{z}} \frac{(2 \tan \sqrt{z} + 1 - \tan^2 \sqrt{z})}{(2 \sqrt{z} \sec^2 \sqrt{z})} \, dz \)
2. \( y = \int_{0}^{t^2} e^{\sqrt{z}} \frac{(1 - \tan^2 \sqrt{z} - 2 \tan \sqrt{z})}{(2 \sqrt{z} \sec^2 \sqrt{z})} \, dz \)

### Step 1: Differentiate \( x \) and \( y \) with respect to \( t \)

Using the Leibniz rule for differentiation under the integral sign:

\[ \frac{dx}{dt} = \frac{d}{dt} \int_{0}^{t^2} f(z) \, dz = f(t^2) \cdot \frac{d(t^2)}{dt} - f(0) \cdot \frac{d(0)}{dt} \]

Since \( f(0) \) will contribute 0, we have:

\[ \frac{dx}{dt} = f(t^2) \cdot 2t \]

Where \( f(z) = e^{\sqrt{z}} \frac{(2 \tan \sqrt{z} + 1 - \tan^2 \sqrt{z})}{(2 \sqrt{z} \sec^2 \sqrt{z})} \).

### Step 2: Evaluate \( f(t^2) \)

Substituting \( z = t^2 \):

\[ f(t^2) = e^{t} \frac{(2 \tan t + 1 - \tan^2 t)}{(2t \sec^2 t)} \]

Thus,

\[ \frac{dx}{dt} = e^{t} \frac{(2 \tan t + 1 - \tan^2 t)}{(2t \sec^2 t)} \cdot 2t = e^{t} (2 \tan t + 1 - \tan^2 t) \cdot \frac{1}{\sec^2 t} \]

### Step 3: Differentiate \( y \)

Similarly, for \( y \):

\[ \frac{dy}{dt} = g(t^2) \cdot 2t \]

Where \( g(z) = e^{\sqrt{z}} \frac{(1 - \tan^2 \sqrt{z} - 2 \tan \sqrt{z})}{(2 \sqrt{z} \sec^2 \sqrt{z})} \).

Thus,

\[ \frac{dy}{dt} = e^{t} \frac{(1 - \tan^2 t - 2 \tan t)}{(2t \sec^2 t)} \cdot 2t = e^{t} (1 - \tan^2 t - 2 \tan t) \cdot \frac{1}{\sec^2 t} \]

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

Now, we can find \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{e^{t} (1 - \tan^2 t - 2 \tan t) \cdot \frac{1}{\sec^2 t}}{e^{t} (2 \tan t + 1 - \tan^2 t) \cdot \frac{1}{\sec^2 t}} = \frac{1 - \tan^2 t - 2 \tan t}{2 \tan t + 1 - \tan^2 t} \]

### Step 5: Evaluate at \( t = \frac{\pi}{4} \)

At \( t = \frac{\pi}{4} \):

\[ \tan \frac{\pi}{4} = 1 \]

Substituting \( t = \frac{\pi}{4} \):

\[ \frac{dy}{dx} = \frac{1 - 1 - 2 \cdot 1}{2 \cdot 1 + 1 - 1} = \frac{-2}{2} = -1 \]

### Step 6: Find the angle of inclination

The angle \( \theta \) of the tangent line with respect to the x-axis is given by:

\[ \tan \theta = \frac{dy}{dx} = -1 \]

Thus,

\[ \theta = \tan^{-1}(-1) = \frac{3\pi}{4} \]

### Conclusion

The inclination of the tangent to the curve at \( t = \frac{\pi}{4} \) is:

\[ \theta = \frac{3\pi}{4} \]