Question

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

Answer 1

We're given to simplify,

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

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,

$\displaystyle\largee^{\log(\tan1\textdegree\tan2\textdegree\tan3\textdegree\dots\ \tan89\textdegree)}$

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,

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

Hence 1 is the answer.

Answer 2

Answer:

answer is zero

Step-by-step explanation:

I hope you understand