Question

\(\log ( \tan 18\degree) + \log ( \tan 36\degree) + \log ( \tan 54\degree) + \log (\tan 72\degree) =\)

• A. $\log 4$
• B. $\log 3$
• C. $\log 2$
• D. 0

Answer 1

The Correct Option is D

$\log ( \tan 18? \tan 36? \tan 54? \tan 72?) = \log 1$ $[\because \:\:If A+ B = 90^\circ \, \Rightarrow \, \tan \, A \, \tan \, B = 1]$

Answer 2

Given:
\(\log(\tan 18^\circ) + \log(\tan 36^\circ) + \log(\tan 54^\circ) + \log(\tan 72^\circ)\)

Combine logarithms using the property \(\log(a) + \log(b) = \log(ab)\):
\(\log(\tan 18^\circ \cdot \tan 36^\circ \cdot \tan 54^\circ \cdot \tan 72^\circ)\)

Use tangent and cotangent identities:
\(\tan(90^\circ - x) = \cot(x)\)
Therefore, \(\tan 72^\circ = \cot 18^\circ\), \(\tan 54^\circ = \cot 36^\circ\), \(\tan 36^\circ = \cot 54^\circ\), \(\tan 18^\circ = \cot 72^\circ\)

Combine all tangents and cotangents:
\(\tan 18^\circ \cdot \tan 36^\circ \cdot \tan 54^\circ \cdot \tan 72^\circ = \tan 18^\circ \cdot \cot 18^\circ \cdot \tan 36^\circ \cdot \cot 36^\circ\)
\(= 1 \cdot 1 = 1\)
The logarithm of 1:
\(\log(1) = 0\)

So, the correct option is (D): 0