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Quantitative Ability and Data Interpretation Question on Logarithms

Sham is trying to solve the expression:
log⁡tan⁡1∘+log⁡tan⁡2∘+log⁡tan⁡3∘+…+log⁡tan⁡89∘.\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

12\frac{1}{\sqrt{2}}

C

0

D

-1

Answer

0

Explanation

Solution

The given expression is:
log⁡tan⁡1∘+log⁡tan⁡2∘+log⁡tan⁡3∘+…+log⁡tan⁡89∘.\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∘+log⁡tan⁡2∘+log⁡tan⁡3∘+…+log⁡tan⁡89∘=log⁡(tan⁡1∘⋅tan⁡2∘⋅tan⁡3∘⋅…⋅tan⁡89∘).\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∘−x)=cot⁡x\tan (90^\circ - x) = \cot x, we can pair the terms:
tan⁡1∘⋅tan⁡89∘=tan⁡1∘⋅cot⁡1∘=1,\tan 1^\circ \cdot \tan 89^\circ = \tan 1^\circ \cdot \cot 1^\circ = 1,
tan⁡2∘⋅tan⁡88∘=tan⁡2∘⋅cot⁡2∘=1,\tan 2^\circ \cdot \tan 88^\circ = \tan 2^\circ \cdot \cot 2^\circ = 1,
tan⁡3∘⋅tan⁡87∘=tan⁡3∘⋅cot⁡3∘=1,\tan 3^\circ \cdot \tan 87^\circ = \tan 3^\circ \cdot \cot 3^\circ = 1,
and so on up to
tan⁡44∘⋅tan⁡46∘=tan⁡44∘⋅cot⁡44∘=1.\tan 44^\circ \cdot \tan 46^\circ = \tan 44^\circ \cdot \cot 44^\circ = 1.
Additionally, tan⁡45∘=1\tan 45^\circ = 1.
Hence, the product of all the terms is:
tan⁡1∘⋅tan⁡2∘⋅tan⁡3∘⋅…⋅tan⁡44∘⋅tan⁡45∘⋅tan⁡46∘⋅…⋅tan⁡89∘=1.\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∘⋅tan⁡2∘⋅tan⁡3∘⋅…⋅tan⁡89∘)=log⁡1=0.\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.