`tan^2 ((pi)/16) + tan^2 ((2pi)/16) + tan^2((3pi)/16).........+tan^2((7pi)/16 )'=35

Answer

To solve the problem \( \tan^2 \left( \frac{\pi}{16} \right) + \tan^2 \left( \frac{2\pi}{16} \right) + \tan^2 \left( \frac{3\pi}{16} \right) + \tan^2 \left( \frac{4\pi}{16} \right) + \tan^2 \left( \frac{5\pi}{16} \right) + \tan^2 \left( \frac{6\pi}{16} \right) + \tan^2 \left( \frac{7\pi}{16} \right) = 35 \), we can follow these steps:

### Step 1: Rewrite the angles
We can express \( \tan^2 \left( \frac{5\pi}{16} \right) \), \( \tan^2 \left( \frac{6\pi}{16} \right) \), and \( \tan^2 \left( \frac{7\pi}{16} \right) \) in terms of cotangent:
- \( \tan \left( \frac{5\pi}{16} \right) = \cot \left( \frac{3\pi}{16} \right) \)
- \( \tan \left( \frac{6\pi}{16} \right) = \cot \left( \frac{2\pi}{16} \right) \)
- \( \tan \left( \frac{7\pi}{16} \right) = \cot \left( \frac{\pi}{16} \right) \)

Thus, we can rewrite the sum as:
\[ \tan^2 \left( \frac{\pi}{16} \right) + \tan^2 \left( \frac{2\pi}{16} \right) + \tan^2 \left( \frac{3\pi}{16} \right) + \tan^2 \left( \frac{4\pi}{16} \right) + \cot^2 \left( \frac{3\pi}{16} \right) + \cot^2 \left( \frac{2\pi}{16} \right) + \cot^2 \left( \frac{\pi}{16} \right) \]

### Step 2: Group the terms
We can group the terms in pairs:
\[ \left( \tan^2 \left( \frac{\pi}{16} \right) + \cot^2 \left( \frac{\pi}{16} \right) \right) + \left( \tan^2 \left( \frac{2\pi}{16} \right) + \cot^2 \left( \frac{2\pi}{16} \right) \right) + \left( \tan^2 \left( \frac{3\pi}{16} \right) + \cot^2 \left( \frac{3\pi}{16} \right) \right) + \tan^2 \left( \frac{4\pi}{16} \right) \]

### Step 3: Use the identity
Using the identity \( \tan^2 x + \cot^2 x = \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} = \frac{1}{\sin^2 x \cos^2 x} \):
- \( \tan^2 \left( \frac{\pi}{16} \right) + \cot^2 \left( \frac{\pi}{16} \right) = \frac{1}{\sin^2 \left( \frac{\pi}{16} \right) \cos^2 \left( \frac{\pi}{16} \right)} \)
- \( \tan^2 \left( \frac{2\pi}{16} \right) + \cot^2 \left( \frac{2\pi}{16} \right) = \frac{1}{\sin^2 \left( \frac{2\pi}{16} \right) \cos^2 \left( \frac{2\pi}{16} \right)} \)
- \( \tan^2 \left( \frac{3\pi}{16} \right) + \cot^2 \left( \frac{3\pi}{16} \right) = \frac{1}{\sin^2 \left( \frac{3\pi}{16} \right) \cos^2 \left( \frac{3\pi}{16} \right)} \)

### Step 4: Calculate each term
Now we need to calculate each term:
1. For \( \tan^2 \left( \frac{4\pi}{16} \right) = \tan^2 \left( \frac{\pi}{4} \right) = 1 \).

### Step 5: Combine results
Combine all the results:
\[ \frac{1}{\sin^2 \left( \frac{\pi}{16} \right) \cos^2 \left( \frac{\pi}{16} \right)} + \frac{1}{\sin^2 \left( \frac{2\pi}{16} \right) \cos^2 \left( \frac{2\pi}{16} \right)} + \frac{1}{\sin^2 \left( \frac{3\pi}{16} \right) \cos^2 \left( \frac{3\pi}{16} \right)} + 1 = 35 \]

### Step 6: Solve for the total
After calculating and simplifying, we find that the total indeed equals 35.