If tangent at any point on the curve `e^y = 1 + x^2` makes an angle `theta` with +ive direction of x-axis, then

Options

A. `|tantheta|gt1`
B. `|tantheta|lt1`
C. `tanthetagt1`
D. `|tantheta|le1`

Answer

To solve the problem, we need to find the relationship between the tangent of the angle θ that the tangent line makes with the positive direction of the x-axis and the given curve \( e^y = 1 + x^2 \).

### Step 1: Find the slope of the tangent line
The slope \( M \) of the tangent line at any point on the curve is given by the derivative \( \frac{dy}{dx} \).

### Step 2: Differentiate the curve
Starting with the equation of the curve:
\[ e^y = 1 + x^2 \]
We differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(e^y) = \frac{d}{dx}(1 + x^2) \]
Using the chain rule on the left side:
\[ e^y \frac{dy}{dx} = 0 + 2x \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{2x}{e^y} \]

### Step 3: Substitute \( e^y \)
From the original equation \( e^y = 1 + x^2 \), we can substitute this into our derivative:
\[ \frac{dy}{dx} = \frac{2x}{1 + x^2} \]

### Step 4: Relate the slope to the angle
The slope \( M \) of the tangent line is also related to the angle \( \theta \) by:
\[ M = \tan(\theta) \]
Thus, we can equate:
\[ \tan(\theta) = \frac{2x}{1 + x^2} \]

### Step 5: Find the modulus of the slope
To find the modulus of the slope:
\[ |M| = \left| \frac{2x}{1 + x^2} \right| \]

### Step 6: Analyze the expression
We need to find the condition for \( |M| \):
\[ |M| = \left| \frac{2x}{1 + x^2} \right| \leq 1 \]
This leads us to the inequality:
\[ \left| \frac{2x}{1 + x^2} \right| \leq 1 \]

### Step 7: Solve the inequality
We can rewrite this as:
\[ |2x| \leq |1 + x^2| \]
Since \( 1 + x^2 \) is always positive, we can simplify this to:
\[ 2|x| \leq 1 + x^2 \]
Rearranging gives:
\[ x^2 - 2|x| + 1 \geq 0 \]
This can be factored as:
\[ (|x| - 1)^2 \geq 0 \]
This inequality holds for all \( x \), meaning the original condition \( |M| \leq 1 \) is satisfied.

### Conclusion
Thus, we conclude that:
\[ | \tan(\theta) | \leq 1 \]