The value of `e^(log_(10) tan 1^(@) + log _(10) tan 2^(@) + log _(10) tan 3^(@) +"...."+ log_(10) tan 89^(@))` is

Options

A. 0
B. e
C. `1// e`
D. none of these

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: Simplifying the Power Expression

We start by simplifying the expression in the exponent:

\[ \log_{10} \tan 1^\circ + \log_{10} \tan 2^\circ + \ldots + \log_{10} \tan 89^\circ \]

Using the property of logarithms that states \(\log_{b} a + \log_{b} c = \log_{b} (a \cdot c)\), we can combine the logarithms:

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

### Step 2: Using Trigonometric Identities

Next, we utilize the identity:

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

This implies that:

\[ \tan 89^\circ = \cot 1^\circ, \quad \tan 88^\circ = \cot 2^\circ, \quad \ldots, \quad \tan 46^\circ = \cot 44^\circ \]

Thus, we can pair the terms:

\[ \tan 1^\circ \tan 89^\circ = \tan 1^\circ \cot 1^\circ = 1 \]
\[ \tan 2^\circ \tan 88^\circ = \tan 2^\circ \cot 2^\circ = 1 \]
\[ \ldots \]
\[ \tan 44^\circ \tan 46^\circ = \tan 44^\circ \cot 44^\circ = 1 \]

### Step 3: Counting the Pairs

There are 44 pairs from \(\tan 1^\circ\) to \(\tan 44^\circ\) and their corresponding cotangent pairs from \(\tan 89^\circ\) to \(\tan 46^\circ\). The middle term, \(\tan 45^\circ\), is equal to 1.

Thus, the product of all these terms is:

\[ \tan 1^\circ \tan 89^\circ \cdot \tan 2^\circ \tan 88^\circ \cdots \tan 44^\circ \tan 46^\circ \cdot \tan 45^\circ = 1^{44} \cdot 1 = 1 \]

### Step 4: Final Calculation

Now substituting back into the logarithm:

\[ \log_{10} (\tan 1^\circ \cdot \tan 2^\circ \cdots \tan 89^\circ) = \log_{10} (1) = 0 \]

So, the expression in the exponent simplifies to:

\[ e^{0} = 1 \]

### Conclusion

The value of the original expression is:

\[ \boxed{1} \]