Find the slope of the normal to the curve :
`y=tan^(2)x+secx" at "x=(pi)/(4)`

Answer

To find the slope of the normal to the curve \( y = \tan^2 x + \sec x \) at \( x = \frac{\pi}{4} \), we will follow these steps:

### Step 1: Differentiate the function
We need to find the derivative \( \frac{dy}{dx} \) of the function \( y = \tan^2 x + \sec x \).

Using the chain rule and the derivatives of trigonometric functions, we have:
\[ \frac{dy}{dx} = \frac{d}{dx}(\tan^2 x) + \frac{d}{dx}(\sec x) \]
The derivative of \( \tan^2 x \) is \( 2\tan x \sec^2 x \) and the derivative of \( \sec x \) is \( \sec x \tan x \). Therefore,
\[ \frac{dy}{dx} = 2\tan x \sec^2 x + \sec x \tan x \]

### Step 2: Evaluate the derivative at \( x = \frac{\pi}{4} \)
Now we will substitute \( x = \frac{\pi}{4} \) into the derivative:
\[ \frac{dy}{dx} \bigg|_{x = \frac{\pi}{4}} = 2\tan\left(\frac{\pi}{4}\right) \sec^2\left(\frac{\pi}{4}\right) + \sec\left(\frac{\pi}{4}\right) \tan\left(\frac{\pi}{4}\right) \]
We know that \( \tan\left(\frac{\pi}{4}\right) = 1 \) and \( \sec\left(\frac{\pi}{4}\right) = \sqrt{2} \). Thus,
\[ \frac{dy}{dx} \bigg|_{x = \frac{\pi}{4}} = 2(1)(\sqrt{2})^2 + (\sqrt{2})(1) = 2(2) + \sqrt{2} = 4 + \sqrt{2} \]

### Step 3: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent line. If \( m_t \) is the slope of the tangent, then the slope of the normal \( m_n \) is given by:
\[ m_n = -\frac{1}{m_t} \]
Substituting \( m_t = 4 + \sqrt{2} \):
\[ m_n = -\frac{1}{4 + \sqrt{2}} \]

### Final Answer
Thus, the slope of the normal to the curve at \( x = \frac{\pi}{4} \) is:
\[ m_n = -\frac{1}{4 + \sqrt{2}} \]

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