If `tan(x/2)=cosec x - sin x` then the value of `tan^2(x/2)` is

Answer

To solve the equation \( \tan\left(\frac{x}{2}\right) = \csc x - \sin x \) and find the value of \( \tan^2\left(\frac{x}{2}\right) \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \tan\left(\frac{x}{2}\right) = \csc x - \sin x \]

### Step 2: Use trigonometric identities
Recall that:
\[ \csc x = \frac{1}{\sin x} \]
Thus, we can rewrite the right-hand side:
\[ \tan\left(\frac{x}{2}\right) = \frac{1}{\sin x} - \sin x \]

### Step 3: Combine the terms
To combine the terms on the right-hand side, we need a common denominator:
\[ \tan\left(\frac{x}{2}\right) = \frac{1 - \sin^2 x}{\sin x} = \frac{\cos^2 x}{\sin x} \]
(using the identity \( 1 - \sin^2 x = \cos^2 x \))

### Step 4: Use the double angle formula for sine
We know that:
\[ \sin x = 2 \tan\left(\frac{x}{2}\right) \cdot \frac{1}{1 + \tan^2\left(\frac{x}{2}\right)} \]
Let \( t = \tan\left(\frac{x}{2}\right) \). Then we can express \( \sin x \) in terms of \( t \):
\[ \sin x = \frac{2t}{1 + t^2} \]

### Step 5: Substitute back into the equation
Substituting \( \sin x \) into the equation gives:
\[ t = \frac{\cos^2 x}{\frac{2t}{1 + t^2}} \]

### Step 6: Express \( \cos^2 x \)
Using the identity \( \cos^2 x = 1 - \sin^2 x \):
\[ \cos^2 x = 1 - \left(\frac{2t}{1 + t^2}\right)^2 = 1 - \frac{4t^2}{(1 + t^2)^2} \]

### Step 7: Simplify the equation
Now substituting \( \cos^2 x \) into our equation:
\[ t = \frac{1 - \frac{4t^2}{(1 + t^2)^2}}{\frac{2t}{1 + t^2}} \]

### Step 8: Cross-multiply and simplify
Cross-multiplying gives:
\[ t \cdot \frac{2t}{1 + t^2} = 1 - \frac{4t^2}{(1 + t^2)^2} \]
This leads to:
\[ 2t^2 = (1 + t^2)(1 - \frac{4t^2}{(1 + t^2)^2}) \]

### Step 9: Solve the resulting equation
After simplifying, we arrive at a polynomial equation in terms of \( t^2 \). Let \( y = t^2 \), leading to:
\[ y^2 + 4y - 1 = 0 \]

### Step 10: Use the quadratic formula
Using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-4 \pm \sqrt{16 + 4}}{2} = \frac{-4 \pm \sqrt{20}}{2} \]
This simplifies to:
\[ y = -2 \pm \sqrt{5} \]

### Final Result
Thus, the value of \( \tan^2\left(\frac{x}{2}\right) \) is:
\[ \tan^2\left(\frac{x}{2}\right) = -2 + \sqrt{5} \quad \text{(taking the positive root)} \]