If `A+B+C=180^(@)`, then `tan^(2)""(A)/(2)+tan^(2)""(B)/(2)+tan^(2)""(C)/(2)=?`

Options

A. 1
B. 0
C. 2
D. 0

Answer

To solve the problem, we need to find the value of

\[ \frac{\tan^2(A/2)}{1} + \frac{\tan^2(B/2)}{1} + \frac{\tan^2(C/2)}{1} \]

given that \( A + B + C = 180^\circ \).

### Step 1: Use the condition \( A + B + C = 180^\circ \)

Since \( A + B + C = 180^\circ \), we can choose values for \( A \), \( B \), and \( C \) that satisfy this equation. A common choice is to set \( A = B = C = 60^\circ \).

### Step 2: Calculate \( \tan(A/2) \)

For \( A = 60^\circ \):

\[ \tan\left(\frac{A}{2}\right) = \tan\left(\frac{60^\circ}{2}\right) = \tan(30^\circ) \]

We know that:

\[ \tan(30^\circ) = \frac{1}{\sqrt{3}} \]

### Step 3: Calculate \( \tan^2(A/2) \)

Now, we calculate \( \tan^2(A/2) \):

\[ \tan^2\left(\frac{A}{2}\right) = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \]

### Step 4: Repeat for \( B \) and \( C \)

Since \( B = 60^\circ \) and \( C = 60^\circ \) as well, we can use the same calculations:

\[ \tan^2\left(\frac{B}{2}\right) = \tan^2(30^\circ) = \frac{1}{3} \]
\[ \tan^2\left(\frac{C}{2}\right) = \tan^2(30^\circ) = \frac{1}{3} \]

### Step 5: Sum the values

Now we sum the values:

\[ \tan^2\left(\frac{A}{2}\right) + \tan^2\left(\frac{B}{2}\right) + \tan^2\left(\frac{C}{2}\right) = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} = 1 \]

### Conclusion

Thus, the final answer is:

\[ \frac{\tan^2(A/2)}{1} + \frac{\tan^2(B/2)}{1} + \frac{\tan^2(C/2)}{1} = 1 \]