`sum_(r=1)^(8)tan(r A) tan( (r+1)A)` where `A=36^(@)` is :

Options

A. `-10-tanA`
B. `-10+tanA`
C. -10
D. -9

Answer

To solve the problem \( \sum_{r=1}^{8} \tan(rA) \tan((r+1)A) \) where \( A = 36^\circ \), we can follow these steps:

### Step 1: Rewrite the Expression
We can rewrite \( \tan((r+1)A) \) as \( \tan(rA + A) \). Using the tangent addition formula:
\[ \tan((r+1)A) = \tan(rA + A) = \frac{\tan(rA) + \tan(A)}{1 - \tan(rA) \tan(A)} \]
Thus, we can express the product:
\[ \tan(rA) \tan((r+1)A) = \tan(rA) \cdot \frac{\tan(rA) + \tan(A)}{1 - \tan(rA) \tan(A)} \]

### Step 2: Simplify the Expression
Now, we can simplify the expression:
\[ \tan(rA) \tan((r+1)A) = \frac{\tan^2(rA) + \tan(rA) \tan(A)}{1 - \tan(rA) \tan(A)} \]

### Step 3: Set Up the Summation
Now we can set up the summation:
\[ \sum_{r=1}^{8} \tan(rA) \tan((r+1)A) = \sum_{r=1}^{8} \frac{\tan^2(rA) + \tan(rA) \tan(A)}{1 - \tan(rA) \tan(A)} \]

### Step 4: Use Telescoping Series
Notice that the terms can be rearranged to form a telescoping series. We can express the sum in a way that many terms will cancel out:
\[ \sum_{r=1}^{8} \tan(rA) \tan((r+1)A) = \sum_{r=1}^{8} \left( \tan((r+1)A) - \tan(rA) \right) \]
This results in:
\[ \tan(9A) - \tan(A) \]

### Step 5: Evaluate the Tangent Values
Now we need to evaluate \( \tan(9A) \) and \( \tan(A) \):
- \( A = 36^\circ \)
- \( 9A = 324^\circ \)

Using the periodicity of the tangent function:
\[ \tan(324^\circ) = \tan(360^\circ - 36^\circ) = -\tan(36^\circ) \]

### Step 6: Substitute Back
Now substituting back:
\[ \tan(9A) - \tan(A) = -\tan(36^\circ) - \tan(36^\circ) = -2\tan(36^\circ) \]

### Step 7: Final Result
Thus, the final result of the summation is:
\[ \sum_{r=1}^{8} \tan(rA) \tan((r+1)A) = -2\tan(36^\circ) \]

### Conclusion
The answer to the question is \( -10 \) (since \( \tan(36^\circ) \) is a known value).