If tan x cot `y= sec alpha` where `alpha` is constant and `alpha in (-(pi)/(2), (pi)/(2))then (d ^(2)y)/(dx ^(2))at ((pi)/(4), (pi)/(4))` equal to:

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to find the second derivative \( \frac{d^2y}{dx^2} \) at the point \( \left( \frac{\pi}{4}, \frac{\pi}{4} \right) \) given the equation \( \tan x \cot y = \sec \alpha \).

### Step 1: Rewrite the equation
Starting with the equation:
\[ \tan x \cot y = \sec \alpha \]
We can express \( \cot y \) as \( \frac{1}{\tan y} \):
\[ \tan x \cdot \frac{1}{\tan y} = \sec \alpha \]
This simplifies to:
\[ \tan x = \sec \alpha \cdot \tan y \]

### Step 2: Differentiate both sides with respect to \( x \)
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(\tan x) = \frac{d}{dx}(\sec \alpha \tan y) \]
Using the derivative of \( \tan x \) which is \( \sec^2 x \):
\[ \sec^2 x = \sec \alpha \cdot \sec^2 y \cdot \frac{dy}{dx} \]

### Step 3: Solve for \( \frac{dy}{dx} \)
Rearranging gives:
\[ \frac{dy}{dx} = \frac{\sec^2 x}{\sec \alpha \cdot \sec^2 y} \]

### Step 4: Evaluate \( \frac{dy}{dx} \) at \( \left( \frac{\pi}{4}, \frac{\pi}{4} \right) \)
At \( x = \frac{\pi}{4} \):
\[ \sec^2\left(\frac{\pi}{4}\right) = 2 \]
Thus:
\[ \frac{dy}{dx} = \frac{2}{\sec \alpha \cdot 2} = \frac{1}{\sec \alpha} \]
Since \( \tan\left(\frac{\pi}{4}\right) = 1 \), we have:
\[ \sec \alpha = \tan x \cdot \tan y = 1 \cdot 1 = 1 \]
Therefore:
\[ \frac{dy}{dx} = 1 \]

### Step 5: Differentiate again to find \( \frac{d^2y}{dx^2} \)
Differentiating \( \sec^2 x = \sec \alpha \cdot \sec^2 y \cdot \frac{dy}{dx} \) again:
Using the product rule:
\[ \frac{d}{dx}(\sec^2 x) = 2 \sec^2 x \tan x \]
And applying the product rule on the right-hand side:
\[ 2 \sec^2 x \tan x = \sec \alpha \left( \sec^2 y \cdot \frac{d^2y}{dx^2} + \sec^2 y \cdot \frac{dy}{dx} \cdot 2 \sec y \tan y \right) \]

### Step 6: Substitute known values at \( \left( \frac{\pi}{4}, \frac{\pi}{4} \right) \)
Substituting \( x = \frac{\pi}{4} \):
\[ 2 \cdot 2 \cdot 1 = 1 \left( 2 \cdot \frac{d^2y}{dx^2} + 1 \cdot 2 \cdot 1 \cdot 1 \right) \]
This simplifies to:
\[ 4 = 2 \frac{d^2y}{dx^2} + 2 \]
Rearranging gives:
\[ 2 \frac{d^2y}{dx^2} = 2 \implies \frac{d^2y}{dx^2} = 1 \]

### Final Answer
Thus, the value of \( \frac{d^2y}{dx^2} \) at \( \left( \frac{\pi}{4}, \frac{\pi}{4} \right) \) is:
\[ \frac{d^2y}{dx^2} = 1 \]