If `tan(x/2)=cosec x-sin x`, then`tan^(2)(x/2)` is equal to

Options

A. `2-sqrt(5)`
B. `sqrt(5)-1`
C. `(9-4sqrt(5))(2+sqrt(5))`
D. `(9+4sqrt(5))(2-sqrt(5))`

Answer

To solve the equation \( \tan\left(\frac{x}{2}\right) = \csc x - \sin x \), we need to find \( \tan^2\left(\frac{x}{2}\right) \). Let's go through the solution step by step.

### Step 1: Substitute \( \tan\left(\frac{x}{2}\right) \)
Let \( t = \tan\left(\frac{x}{2}\right) \).

### Step 2: Use the double angle formula for sine
We know that:
\[ \sin x = \frac{2t}{1 + t^2} \]
and since \( \csc x = \frac{1}{\sin x} \), we can write:
\[ \csc x = \frac{1 + t^2}{2t} \]

### Step 3: Substitute into the original equation
Now we substitute these into the original equation:
\[ t = \csc x - \sin x \]
This gives us:
\[ t = \frac{1 + t^2}{2t} - \frac{2t}{1 + t^2} \]

### Step 4: Find a common denominator
The common denominator for the right-hand side is \( 2t(1 + t^2) \). Thus, we rewrite the equation:
\[ t = \frac{(1 + t^2)^2 - 4t^2}{2t(1 + t^2)} \]

### Step 5: Cross-multiply
Cross-multiplying gives us:
\[ 2t^2(1 + t^2) = (1 + t^2)^2 - 4t^2 \]

### Step 6: Expand and simplify
Expanding both sides:
\[ 2t^2 + 2t^4 = 1 + 2t^2 + t^4 - 4t^2 \]
This simplifies to:
\[ 2t^4 + 2t^2 = 1 - 2t^2 + t^4 \]
Rearranging gives:
\[ t^4 + 4t^2 - 1 = 0 \]

### Step 7: Solve the quadratic equation
Let \( u = t^2 \). The equation becomes:
\[ u^2 + 4u - 1 = 0 \]
Using the quadratic formula:
\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 + 4}}{2} = \frac{-4 \pm \sqrt{20}}{2} = -2 \pm \sqrt{5} \]

### Step 8: Determine the valid solution
Since \( t^2 \) must be non-negative, we take:
\[ t^2 = -2 + \sqrt{5} \]

### Step 9: Conclusion
Thus, we find:
\[ \tan^2\left(\frac{x}{2}\right) = -2 + \sqrt{5} \]

### Final Answer
\[ \tan^2\left(\frac{x}{2}\right) = -2 + \sqrt{5} \]

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