If `y=tan x. cos^(2)x` then `(dy)/(dx)` will be-

Options

A. `1+2sin^(2)x`
B. `1-2sin^(2)x`
C. `1`
D. `2 sin^(2)x`

Explanation

To solve the problem of finding the derivative of \( y = \tan x \cdot \cos^2 x \), we will follow these steps:

### Step 1: Rewrite the function in a simpler form
We know that \( \tan x = \frac{\sin x}{\cos x} \). Therefore, we can rewrite \( y \) as:
\[ y = \tan x \cdot \cos^2 x = \frac{\sin x}{\cos x} \cdot \cos^2 x = \sin x \cdot \cos x \]

### Step 2: Differentiate using the product rule
Now we need to differentiate \( y = \sin x \cdot \cos x \). We will use the product rule, which states that if \( y = u \cdot v \), then:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
Here, let \( u = \sin x \) and \( v = \cos x \).

### Step 3: Find the derivatives of \( u \) and \( v \)
We know:
\[ \frac{du}{dx} = \cos x \quad \text{and} \quad \frac{dv}{dx} = -\sin x \]

### Step 4: Apply the product rule
Now substituting into the product rule:
\[ \frac{dy}{dx} = \sin x \cdot (-\sin x) + \cos x \cdot \cos x \]
This simplifies to:
\[ \frac{dy}{dx} = -\sin^2 x + \cos^2 x \]

### Step 5: Use the Pythagorean identity
Recall that \( \cos^2 x = 1 - \sin^2 x \). Substituting this into our derivative gives:
\[ \frac{dy}{dx} = -\sin^2 x + (1 - \sin^2 x) \]
This simplifies to:
\[ \frac{dy}{dx} = 1 - 2\sin^2 x \]

### Final Answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = 1 - 2\sin^2 x \]