If `(sinx)/(1+cosx)+(sinx)/(1-cosx)=4` and `0^@ le x le 90^@`, then the value of x is

Options

A. `10^@`
B. `15^@`
C. `30^@`
D. `45^@`

Answer

To solve the equation

\[ \frac{\sin x}{1 + \cos x} + \frac{\sin x}{1 - \cos x} = 4 \]

for \(0^\circ \leq x \leq 90^\circ\), we can follow these steps:

### Step 1: Combine the fractions

We can find a common denominator for the left-hand side:

\[ \frac{\sin x (1 - \cos x) + \sin x (1 + \cos x)}{(1 + \cos x)(1 - \cos x)} = 4 \]

### Step 2: Simplify the numerator

The numerator simplifies as follows:

\[ \sin x (1 - \cos x) + \sin x (1 + \cos x) = \sin x - \sin x \cos x + \sin x + \sin x \cos x = 2\sin x \]

So, we rewrite the equation as:

\[ \frac{2\sin x}{(1 + \cos x)(1 - \cos x)} = 4 \]

### Step 3: Simplify the denominator

The denominator can be simplified using the identity \(1 - \cos^2 x = \sin^2 x\):

\[ (1 + \cos x)(1 - \cos x) = 1 - \cos^2 x = \sin^2 x \]

Thus, the equation becomes:

\[ \frac{2\sin x}{\sin^2 x} = 4 \]

### Step 4: Cross-multiply

Cross-multiplying gives:

\[ 2\sin x = 4\sin^2 x \]

### Step 5: Rearrange the equation

Rearranging this gives:

\[ 4\sin^2 x - 2\sin x = 0 \]

### Step 6: Factor the equation

Factoring out \(2\sin x\):

\[ 2\sin x(2\sin x - 1) = 0 \]

### Step 7: Solve for \(\sin x\)

Setting each factor to zero gives us:

1. \(2\sin x = 0\) → \(\sin x = 0\) (not in the range \(0^\circ \leq x \leq 90^\circ\))
2. \(2\sin x - 1 = 0\) → \(\sin x = \frac{1}{2}\)

### Step 8: Find the angle \(x\)

The angle \(x\) for which \(\sin x = \frac{1}{2}\) in the range \(0^\circ \leq x \leq 90^\circ\) is:

\[ x = 30^\circ \]

### Final Answer

Thus, the value of \(x\) is:

\[ \boxed{30^\circ} \]