The value of the expression `log_(10) (tan 6^(@))+log_(10) (tan 12^(@))+log_(10) (tan 18^(@))+...+log_(10)(tan 84^(@))` is -

Options

A. a whole number
B. an irrational number
C. a negative integer
D. a rationa number which is not an integer

Answer

To solve the expression

\[ \log_{10}(\tan 6^\circ) + \log_{10}(\tan 12^\circ) + \log_{10}(\tan 18^\circ) + \ldots + \log_{10}(\tan 84^\circ), \]

we can use the properties of logarithms and trigonometric identities.

### Step 1: Use the property of logarithms
Using the property of logarithms that states \(\log_a b + \log_a c = \log_a(b \cdot c)\), we can combine the logarithmic terms:

\[ \log_{10}(\tan 6^\circ \cdot \tan 12^\circ \cdot \tan 18^\circ \cdots \tan 84^\circ). \]

### Step 2: Identify the pairs of angles
Next, we can pair the terms in the product. Notice that:

\[ \tan(90^\circ - \theta) = \cot(\theta). \]

Thus, we can pair the angles:

- \(\tan 6^\circ\) pairs with \(\tan 84^\circ\) (since \(\tan 84^\circ = \cot 6^\circ\)),
- \(\tan 12^\circ\) pairs with \(\tan 78^\circ\) (since \(\tan 78^\circ = \cot 12^\circ\)),
- \(\tan 18^\circ\) pairs with \(\tan 72^\circ\) (since \(\tan 72^\circ = \cot 18^\circ\)),
- \(\tan 24^\circ\) pairs with \(\tan 66^\circ\) (since \(\tan 66^\circ = \cot 24^\circ\)),
- \(\tan 30^\circ\) pairs with \(\tan 60^\circ\) (since \(\tan 60^\circ = \cot 30^\circ\)),
- \(\tan 36^\circ\) pairs with \(\tan 54^\circ\) (since \(\tan 54^\circ = \cot 36^\circ\)),
- \(\tan 42^\circ\) pairs with \(\tan 48^\circ\) (since \(\tan 48^\circ = \cot 42^\circ\)).

### Step 3: Simplify the product
Each pair of terms can be simplified:

\[ \tan \theta \cdot \cot \theta = 1. \]

Thus, each of the pairs contributes a product of 1:

\[ \tan 6^\circ \cdot \tan 84^\circ = 1, \]
\[ \tan 12^\circ \cdot \tan 78^\circ = 1, \]
\[ \tan 18^\circ \cdot \tan 72^\circ = 1, \]
\[ \tan 24^\circ \cdot \tan 66^\circ = 1, \]
\[ \tan 30^\circ \cdot \tan 60^\circ = 1, \]
\[ \tan 36^\circ \cdot \tan 54^\circ = 1, \]
\[ \tan 42^\circ \cdot \tan 48^\circ = 1. \]

### Step 4: Count the pairs
There are 7 pairs, and since each pair equals 1, the product of all terms is:

\[ 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 \cdot 1 = 1. \]

### Step 5: Apply the logarithm
Now we substitute back into the logarithm:

\[ \log_{10}(1) = 0. \]

### Conclusion
Thus, the value of the original expression is:

\[ \boxed{0}. \]