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 `y=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 problem step by step, we need to find the inclination of the tangent to the curve defined by the integrals \( x \) and \( y \) at \( t = \frac{\pi}{4} \).

### Step 1: Define the integrals
We have:
\[ x = \int_{0}^{t^2} e^{\sqrt{z}} \frac{2 \tan(\sqrt{z}) + 1 - \tan^2(\sqrt{z})}{2 \sqrt{z} \sec^2(\sqrt{z})} \, dz \]
\[ y = \int_{0}^{t^2} e^{\sqrt{z}} \frac{1 - \tan^2(\sqrt{z}) - 2 \tan(\sqrt{z})}{2 \sqrt{z} \sec^2(\sqrt{z})} \, dz \]

### Step 2: Differentiate \( x \) and \( y \) with respect to \( t \)
Using the Fundamental Theorem of Calculus and the chain rule, we differentiate \( x \) and \( y \):
\[ \frac{dx}{dt} = \frac{d}{dt} \left( \int_{0}^{t^2} e^{\sqrt{z}} \frac{2 \tan(\sqrt{z}) + 1 - \tan^2(\sqrt{z})}{2 \sqrt{z} \sec^2(\sqrt{z})} \, dz \right) = 2t e^{t} \frac{2 \tan(t) + 1 - \tan^2(t)}{2t \sec^2(t)} \]
\[ \frac{dy}{dt} = \frac{d}{dt} \left( \int_{0}^{t^2} e^{\sqrt{z}} \frac{1 - \tan^2(\sqrt{z}) - 2 \tan(\sqrt{z})}{2 \sqrt{z} \sec^2(\sqrt{z})} \, dz \right) = 2t e^{t} \frac{1 - \tan^2(t) - 2 \tan(t)}{2t \sec^2(t)} \]

### Step 3: Simplify \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \)
Both derivatives can be simplified:
\[ \frac{dx}{dt} = e^{t} \frac{2 \tan(t) + 1 - \tan^2(t)}{\sec^2(t)} \]
\[ \frac{dy}{dt} = e^{t} \frac{1 - \tan^2(t) - 2 \tan(t)}{\sec^2(t)} \]

### Step 4: Find \( \frac{dy}{dx} \)
Using the chain rule:
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{1 - \tan^2(t) - 2 \tan(t)}{2 \tan(t) + 1 - \tan^2(t)} \]

### Step 5: Evaluate \( \frac{dy}{dx} \) at \( t = \frac{\pi}{4} \)
At \( t = \frac{\pi}{4} \):
\[ \tan\left(\frac{\pi}{4}\right) = 1 \]
Substituting \( t = \frac{\pi}{4} \):
\[ \frac{dy}{dx} = \frac{1 - 1 - 2(1)}{2(1) + 1 - 1} = \frac{-2}{2} = -1 \]

### Step 6: Find the inclination of the tangent
The inclination of the tangent is given by the angle whose tangent is \( \frac{dy}{dx} \):
\[ \theta = \tan^{-1}(-1) \]
This corresponds to an angle of \( \frac{3\pi}{4} \) (since the tangent is negative in the second quadrant).

### Final Answer
Thus, the inclination of the tangent to the curve at \( t = \frac{\pi}{4} \) is:
\[ \frac{3\pi}{4} \]