If ` y= sqrt ((1+sin x) /( 1-sin x) ,)then (dy)/(dx) =`

Options

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

Answer

To find the derivative of the function \( y = \sqrt{\frac{1 + \sin x}{1 - \sin x}} \), we will follow these steps:

### Step 1: Rationalize the expression
We start with the expression for \( y \):
\[ y = \sqrt{\frac{1 + \sin x}{1 - \sin x}} \]
To simplify this, we can multiply the numerator and denominator by \( 1 + \sin x \):
\[ y = \sqrt{\frac{(1 + \sin x)(1 + \sin x)}{(1 - \sin x)(1 + \sin x)}} \]
This simplifies to:
\[ y = \sqrt{\frac{(1 + \sin x)^2}{1 - \sin^2 x}} \]

### Step 2: Simplify using trigonometric identities
Using the identity \( 1 - \sin^2 x = \cos^2 x \), we can rewrite the expression:
\[ y = \sqrt{\frac{(1 + \sin x)^2}{\cos^2 x}} \]
Now, we can take the square root:
\[ y = \frac{1 + \sin x}{\cos x} \]

### Step 3: Rewrite the expression
We can separate the terms:
\[ y = \frac{1}{\cos x} + \frac{\sin x}{\cos x} \]
This simplifies to:
\[ y = \sec x + \tan x \]

### Step 4: Differentiate the expression
Now, we differentiate \( y \):
\[ \frac{dy}{dx} = \frac{d}{dx}(\sec x) + \frac{d}{dx}(\tan x) \]
Using the derivatives \( \frac{d}{dx}(\sec x) = \sec x \tan x \) and \( \frac{d}{dx}(\tan x) = \sec^2 x \), we get:
\[ \frac{dy}{dx} = \sec x \tan x + \sec^2 x \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \sec x \tan x + \sec^2 x \]

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