Question

Sham is trying to solve the expression:
\[ \log \tan 1^\circ + \log \tan 2^\circ + \log \tan 3^\circ + \ldots + \log \tan 89^\circ. \]  
The correct answer would be?

• A. 1
• B. \(\frac{1}{\sqrt{2}}\)
• C. 0
• D. -1

Answer 1

The Correct Option is C

The given expression is:
\[ \log \tan 1^\circ + \log \tan 2^\circ + \log \tan 3^\circ + \ldots + \log \tan 89^\circ. \]
We can rewrite the expression using the properties of logarithms:  
\[ \log \tan 1^\circ + \log \tan 2^\circ + \log \tan 3^\circ + \ldots + \log \tan 89^\circ = \log (\tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \cdot \ldots \cdot \tan 89^\circ). \]
Using the identity \(\tan (90^\circ - x) = \cot x\), we can pair the terms:
\[ \tan 1^\circ \cdot \tan 89^\circ = \tan 1^\circ \cdot \cot 1^\circ = 1, \]
\[ \tan 2^\circ \cdot \tan 88^\circ = \tan 2^\circ \cdot \cot 2^\circ = 1, \]
\[ \tan 3^\circ \cdot \tan 87^\circ = \tan 3^\circ \cdot \cot 3^\circ = 1, \]
and so on up to
\[ \tan 44^\circ \cdot \tan 46^\circ = \tan 44^\circ \cdot \cot 44^\circ = 1. \]
Additionally, \(\tan 45^\circ = 1\).
Hence, the product of all the terms is:
\[ \tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \cdot \ldots \cdot \tan 44^\circ \cdot \tan 45^\circ \cdot \tan 46^\circ \cdot \ldots \cdot \tan 89^\circ = 1. \]
Therefore, the logarithm of the product is:
\[ \log (\tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \cdot \ldots \cdot \tan 89^\circ) = \log 1 = 0. \]
Thus, the correct answer is Option C :0.