If `(1+ sintheta-costheta)/(1+ sintheta + costheta)+(1+sintheta+costheta)/(1+sintheta-costheta)= 4`, then which of the following values will be suitable for `theta` ?

Options

A. A) `45^@`
B. B) `30^@`
C. C) `90^@`
D. D) `60^@`

Answer

To solve the equation

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

we will simplify the left-hand side step by step.

### Step 1: Simplify the expression
Let \( A = 1 + \sin \theta \) and \( B = \cos \theta \). The equation can be rewritten as:

\[ \frac{A - B}{A + B} + \frac{A + B}{A - B} = 4. \]

### Step 2: Find a common denominator
The common denominator for the two fractions is \( (A + B)(A - B) \). Thus, we can write:

\[ \frac{(A - B)^2 + (A + B)^2}{(A + B)(A - B)} = 4. \]

### Step 3: Expand the numerator
Now, we expand the numerator:

\[ (A - B)^2 + (A + B)^2 = (A^2 - 2AB + B^2) + (A^2 + 2AB + B^2) = 2A^2 + 2B^2. \]

### Step 4: Substitute back
Substituting back, we have:

\[ \frac{2(A^2 + B^2)}{(A + B)(A - B)} = 4. \]

### Step 5: Cross-multiply
Cross-multiplying gives:

\[ 2(A^2 + B^2) = 4(A + B)(A - B). \]

### Step 6: Simplify further
This simplifies to:

\[ A^2 + B^2 = 2(A + B)(A - B). \]

### Step 7: Substitute \( A \) and \( B \)
Now substituting back \( A = 1 + \sin \theta \) and \( B = \cos \theta \):

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

### Step 8: Use trigonometric identities
Using the identity \( \sin^2 \theta + \cos^2 \theta = 1 \):

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

### Step 9: Solve for specific values of \( \theta \)
Now we can test specific values of \( \theta \) to see if they satisfy the original equation.

1. **Testing \( \theta = 90^\circ \)**:
- \( \sin 90^\circ = 1 \) and \( \cos 90^\circ = 0 \).
- Substitute:
\[ \frac{1 + 1 - 0}{1 + 1 + 0} + \frac{1 + 1 + 0}{1 + 1 - 0} = \frac{2}{2} + \frac{2}{2} = 1 + 1 = 2 \quad (\text{not } 4) \]

2. **Testing \( \theta = 30^\circ \)**:
- \( \sin 30^\circ = \frac{1}{2} \) and \( \cos 30^\circ = \frac{\sqrt{3}}{2} \).
- Substitute:
\[ \frac{1 + \frac{1}{2} - \frac{\sqrt{3}}{2}}{1 + \frac{1}{2} + \frac{\sqrt{3}}{2}} + \frac{1 + \frac{1}{2} + \frac{\sqrt{3}}{2}}{1 + \frac{1}{2} - \frac{\sqrt{3}}{2}}. \]

After simplifying, you will find that this does indeed equal 4.

### Conclusion
The suitable value for \( \theta \) is \( 30^\circ \).