The value of `log_(3) tan 1^(@) + log_(3) tan 2^(@) + ….+ log_(3) tan 89^(@)` is

Options

A. 3
B. 1
C. 2
D. 0

Answer

To solve the problem \( \log_{3} \tan 1^{\circ} + \log_{3} \tan 2^{\circ} + \ldots + \log_{3} \tan 89^{\circ} \), we can use properties of logarithms and trigonometric identities. Here’s a step-by-step solution:

### Step 1: Use the property of logarithms
We know that \( \log_{a} b + \log_{a} c = \log_{a} (bc) \). Therefore, we can combine the logarithmic terms:

\[ \log_{3} \tan 1^{\circ} + \log_{3} \tan 2^{\circ} + \ldots + \log_{3} \tan 89^{\circ} = \log_{3} (\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}) \]

### Step 2: Use the identity of tangent
Using the identity \( \tan(90^{\circ} - x) = \cot x \), we can pair the terms:

\[ \tan 1^{\circ} \tan 89^{\circ}, \tan 2^{\circ} \tan 88^{\circ}, \ldots, \tan 44^{\circ} \tan 46^{\circ} \]

Each pair can be expressed as:

\[ \tan k^{\circ} \tan (90^{\circ} - k^{\circ}) = \tan k^{\circ} \cot k^{\circ} = 1 \]

### Step 3: Count the pairs
Since we have pairs from \( 1^{\circ} \) to \( 44^{\circ} \) and the middle term \( \tan 45^{\circ} = 1 \):

- The pairs are \( (1^{\circ}, 89^{\circ}), (2^{\circ}, 88^{\circ}), \ldots, (44^{\circ}, 46^{\circ}) \), which gives us 44 pairs.
- The middle term is \( \tan 45^{\circ} = 1 \).

Thus, we have:

\[ \tan 1^{\circ} \tan 89^{\circ} \cdots \tan 44^{\circ} \tan 46^{\circ} \tan 45^{\circ} = 1^{44} \cdot 1 = 1 \]

### Step 4: Substitute back into logarithm
Now substituting back into our logarithmic expression:

\[ \log_{3} (\tan 1^{\circ} \tan 2^{\circ} \ldots \tan 89^{\circ}) = \log_{3} (1) = 0 \]

### Final Answer
Thus, the value of \( \log_{3} \tan 1^{\circ} + \log_{3} \tan 2^{\circ} + \ldots + \log_{3} \tan 89^{\circ} \) is:

\[ \boxed{0} \]