If `tan ""A/2 =3/2, ` then `(1+ cos A)/(1- cos A)=`

Options

A. `-5`
B. 5
C. `9/4`
D. `4/9`

Answer

To solve the problem, we need to find the value of \(\frac{1 + \cos A}{1 - \cos A}\) given that \(\tan \frac{A}{2} = \frac{3}{2}\).

### Step 1: Use the half-angle identity for cosine
We can use the identity for \(\cos A\) in terms of \(\tan \frac{A}{2}\):
\[ \cos A = \frac{1 - \tan^2 \frac{A}{2}}{1 + \tan^2 \frac{A}{2}} \]

### Step 2: Substitute the value of \(\tan \frac{A}{2}\)
Given that \(\tan \frac{A}{2} = \frac{3}{2}\), we can substitute this into the formula:
\[ \cos A = \frac{1 - \left(\frac{3}{2}\right)^2}{1 + \left(\frac{3}{2}\right)^2} \]

### Step 3: Calculate \(\tan^2 \frac{A}{2}\)
Calculate \(\tan^2 \frac{A}{2}\):
\[ \tan^2 \frac{A}{2} = \left(\frac{3}{2}\right)^2 = \frac{9}{4} \]

### Step 4: Substitute \(\tan^2 \frac{A}{2}\) into the cosine formula
Now substitute \(\tan^2 \frac{A}{2}\) into the expression for \(\cos A\):
\[ \cos A = \frac{1 - \frac{9}{4}}{1 + \frac{9}{4}} = \frac{\frac{4}{4} - \frac{9}{4}}{\frac{4}{4} + \frac{9}{4}} = \frac{-\frac{5}{4}}{\frac{13}{4}} = -\frac{5}{13} \]

### Step 5: Substitute \(\cos A\) into \(\frac{1 + \cos A}{1 - \cos A}\)
Now we can substitute \(\cos A\) into the expression \(\frac{1 + \cos A}{1 - \cos A}\):
\[ \frac{1 + \cos A}{1 - \cos A} = \frac{1 - \frac{5}{13}}{1 + \frac{5}{13}} = \frac{\frac{13}{13} - \frac{5}{13}}{\frac{13}{13} + \frac{5}{13}} = \frac{\frac{8}{13}}{\frac{18}{13}} = \frac{8}{18} = \frac{4}{9} \]

### Final Answer
Thus, the value of \(\frac{1 + \cos A}{1 - \cos A}\) is \(\frac{4}{9}\).

---