Question

The value of \(e^{\log_{10}\tan 1°+\log_{10}\tan 2°+\log_{10}\tan 3°+...+\log_{10}\tan89°}\) is

• A. \(\frac{1}{e}\)
• B. 0
• C. 1
• D. 3

Hint:

When dealing with sums of logarithms, use the logarithmic identity \( \log_b x + \log_b y = \log_b (x \cdot y) \) to combine the terms. Also, recognize symmetries in trigonometric functions (like \( \tan(90^\circ - x) = \cot x \)) to simplify the product of tangents.

Answer 1

The Correct Option is C

The correct answer is: (C): 1

We are tasked with finding the value of the following expression:

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

Step 1: Simplify the sum of logarithms

The given expression involves a sum of logarithms. Using the logarithmic identity \[ \log_b x + \log_b y = \log_b (x \cdot y) \], we can combine all the logarithms into one:

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

Step 2: Use the symmetry of the tangent function

We know that \[ \tan(90^\circ - x) = \cot x \], so each pair of terms \( \tan x \) and \( \tan(90^\circ - x) \) will multiply to 1:

\[ \tan 1^\circ \cdot \tan 89^\circ = 1, \quad \tan 2^\circ \cdot \tan 88^\circ = 1, \quad \ldots, \quad \tan 44^\circ \cdot \tan 46^\circ = 1, \quad \tan 45^\circ = 1 \]

Thus, the entire product simplifies to 1:

\[ \tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \cdots \tan 89^\circ = 1 \]

Step 3: Simplify the expression

Now substitute this result back into the original expression:

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

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

\[ e^0 = 1 \]

Conclusion:
The value of the expression is 1, so the correct answer is (C): 1.