the value of `e^(log_(10)tan1^@+log_(10)tan2^@+log_(10)tan3^@....+log_(10)tan89^@`

Options

A. 0
B. e
C. `1//e`
D. 1

Answer

To solve the problem, we need to evaluate the expression:

\[ e^{\log_{10}(\tan 1^\circ) + \log_{10}(\tan 2^\circ) + \log_{10}(\tan 3^\circ) + \ldots + \log_{10}(\tan 89^\circ)} \]

### Step 1: Use the property of logarithms

We know that the sum of logarithms can be combined into a single logarithm:

\[ \log_{10}(a) + \log_{10}(b) = \log_{10}(a \cdot b) \]

Applying this property, we can rewrite our expression as:

\[ e^{\log_{10}(\tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \cdots \tan 89^\circ)} \]

### Step 2: Simplify the expression inside the logarithm

Next, we need to evaluate the product \(\tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \cdots \tan 89^\circ\).

Using the identity \(\tan(90^\circ - \theta) = \cot(\theta)\), we can pair terms:

\[ \tan 1^\circ \cdot \tan 89^\circ = \tan 1^\circ \cdot \cot 1^\circ = 1 \]
\[ \tan 2^\circ \cdot \tan 88^\circ = \tan 2^\circ \cdot \cot 2^\circ = 1 \]
\[ \tan 3^\circ \cdot \tan 87^\circ = \tan 3^\circ \cdot \cot 3^\circ = 1 \]
\[ \vdots \]
\[ \tan 44^\circ \cdot \tan 46^\circ = \tan 44^\circ \cdot \cot 44^\circ = 1 \]

The middle term, \(\tan 45^\circ\), equals 1. Thus, the entire product simplifies to:

\[ \tan 1^\circ \cdot \tan 2^\circ \cdots \tan 44^\circ \cdot \tan 45^\circ \cdot \tan 46^\circ \cdots \tan 89^\circ = 1 \]

### Step 3: Substitute back into the logarithm

Now we can substitute this result back into our expression:

\[ e^{\log_{10}(1)} \]

### Step 4: Evaluate the logarithm

Since \(\log_{10}(1) = 0\), we have:

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

### Step 5: Final result

Finally, we know that \(e^0 = 1\). Therefore, the value of the original expression is:

\[ \boxed{1} \]

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