Math, asked by kalitanitali7797, 1 year ago

E^log10 tan1°+log10 tan2°+log10 tan3°+.......+log10 tan89°=?

Answers

Answered by shadowsabers03
6

We're given to simplify,

\displaystyle\large\text{$e^{\log(\tan1\textdegree)+\log(\tan2\textdegree)+\log(\tan3\textdegree)+\dots\ +\log(\tan89\textdegree)}$}

Since the base is same and 10 in each logarithm mentioned in the exponent, there's no need to give the base.

We are familiar with the logarithmic identity,

\log a_1+\log a_2+\log a_3+\dots\ +\log a_n=\log\left(a_1\cdot a_2\cdot a_3\cdot \dots\dots\ \cdot a_n\right)

Or simply,  \displaystyle\sum_{k=1}^{n}\log a_k\ =\ \log\left(\prod_{k=1}^{n}a_k\right)

So we get,

\displaystyle\large\text{$e^{\log(\tan1\textdegree\tan2\textdegree\tan3\textdegree\dots\ \tan89\textdegree)}$}

But what about  \tan1\textdegree\tan2\textdegree\tan3\textdegree\dots\ \tan89\textdegree\ ?

\tan1\textdegree\tan2\textdegree\tan3\textdegree\dots\ \tan44\textdegree\tan45\textdegree\tan46\textdegree\dots\ \tan87\textdegree\tan88\textdegree\tan89\textdegree\\\\=(\tan1\textdegree\tan89\textdegree)\cdot(\tan2\textdegree\tan88\textdegree)\cdot(\tan3\textdegree\tan87\textdegree)\dots\ (\tan44\textdegree\tan46\textdegree)\cdot\tan45\textdegree

Since \tan x=\tan(90\textdegree-x),

=(\tan1\textdegree\cot1\textdegree)\cdot(\tan2\textdegree\cot2\textdegree)\cdot(\tan3\textdegree\cot3\textdegree)\dots\ (\tan44\textdegree\cot44\textdegree)\cdot1\\\\=1

Because  \tan x\cdot\cot x=1.

Hence,

\displaystyle\large\text{$e^{\log(\tan1\textdegree\tan2\textdegree\tan3\textdegree\dots\ \tan89\textdegree)}$}=e^{\log1}=e^0=\mathbf{1}

Hence 1 is the answer.

Answered by srinivasaraothota000
1

Answer:

answer is zero

Step-by-step explanation:

I hope you understand

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