If sec x + tan x =22/7 , find the value of tanx/2

Options

A. the value of tan `(x)/(2)=(29)/(15)`
B. the value of tan `(x)/(2)=(29)/(15)`
C. the value of cosec x + cot `x=(29)/(15)`
D. thevalue of cosec x + cot `x=(15)/(29)`

Explanation

To solve the problem where \( \sec x + \tan x = \frac{22}{7} \) and we need to find the value of \( \tan \frac{x}{2} \), we can follow these steps:

### Step 1: Rewrite the equation
We know that:
\[ \sec x = \frac{1}{\cos x} \quad \text{and} \quad \tan x = \frac{\sin x}{\cos x} \]
Thus, we can rewrite the equation as:
\[ \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{22}{7} \]
Combining the fractions gives:
\[ \frac{1 + \sin x}{\cos x} = \frac{22}{7} \]

### Step 2: Cross-multiply
Cross-multiplying gives:
\[ 7(1 + \sin x) = 22 \cos x \]
This simplifies to:
\[ 7 + 7 \sin x = 22 \cos x \]

### Step 3: Use the half-angle formulas
We can express \( \sin x \) and \( \cos x \) in terms of \( \tan \frac{x}{2} \):
\[ \sin x = \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \quad \text{and} \quad \cos x = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \]

### Step 4: Substitute into the equation
Substituting these into our equation gives:
\[ 7 + 7 \left(\frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}\right) = 22 \left(\frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}\right) \]

### Step 5: Clear the denominators
Multiply through by \( 1 + \tan^2 \frac{x}{2} \) to eliminate the fractions:
\[ 7(1 + \tan^2 \frac{x}{2}) + 14 \tan \frac{x}{2} = 22(1 - \tan^2 \frac{x}{2}) \]

### Step 6: Rearrange the equation
Expanding and rearranging gives:
\[ 7 + 7 \tan^2 \frac{x}{2} + 14 \tan \frac{x}{2} = 22 - 22 \tan^2 \frac{x}{2} \]
Combining like terms results in:
\[ 29 \tan^2 \frac{x}{2} + 14 \tan \frac{x}{2} - 15 = 0 \]

### Step 7: Solve the quadratic equation
Let \( y = \tan \frac{x}{2} \). The equation becomes:
\[ 29y^2 + 14y - 15 = 0 \]
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ y = \frac{-14 \pm \sqrt{14^2 - 4 \cdot 29 \cdot (-15)}}{2 \cdot 29} \]
Calculating the discriminant:
\[ 14^2 = 196 \quad \text{and} \quad 4 \cdot 29 \cdot 15 = 1740 \]
Thus,
\[ b^2 - 4ac = 196 + 1740 = 1936 \]
Taking the square root:
\[ \sqrt{1936} = 44 \]
Now substituting back:
\[ y = \frac{-14 \pm 44}{58} \]
Calculating the two possible values:
1. \( y = \frac{30}{58} = \frac{15}{29} \)
2. \( y = \frac{-58}{58} = -1 \) (not valid since \( \tan \frac{x}{2} \) cannot be negative)

### Final Answer
Thus, the value of \( \tan \frac{x}{2} \) is:
\[ \boxed{\frac{15}{29}} \]