Log of 89 Tangents

log(tanx) + log(tany) = 0 [where x , y both are integer & ( x + y = 90 )] and log(tan45) = 0 ; now try yourself .

We should know the formulas $\tan (90^{\circ}-A^{\circ})=\cot A^{\circ}$ , $\log P + \log Q = \log (P\times Q)$ . Also, you must know that $\tan \theta \times \cot \theta = 1$ , $\tan 45^{\circ} = 1$ and $\log 1=0$ . Now ----

$\log (\tan 1^{\circ})+\log (\tan 2^{\circ})+ ........ +\log (\tan 89^{\circ})$

$=\log (\tan 1^{\circ})+\log (\tan 2^{\circ})+ ..... +\log (\tan 45^{\circ})+\log (\tan (90^{\circ}-44^{\circ}))+\log (\tan (90^{\circ}-43^{\circ}))+ ....... +\log (\tan (90^{\circ}-1^{\circ}))$

$=\log (\tan 1^{\circ})+\log (\tan 2^{\circ})+ ..... +\log 1 +\log (\cot 44^{\circ})+\log (\cot 43^{\circ})+ ...... +\log (\cot 1^{\circ})$

$=\log (\tan 1^{\circ}\times \cot 1^{\circ})+\log (\tan 2^{\circ}\times \cot 2^{\circ})+ ...... +\log (\tan 44^{\circ}\times \cot 44^{\circ})+\log 1$

$=\log 1 + \log 1 + ........ + \log 1 = 0+0+ ..... +0 = \boxed{0}$

I think you meant $\displaystyle \log (\tan 1^\circ)+\log (\tan 2^\circ)+\ldots+\log (\tan 89^\circ)$ .

Anyway, apply the logarithm sum rule $\displaystyle \log a + \log b = \log (ab)$ to find $\displaystyle \log (\tan 1^\circ*tan 2^\circ*\ldots*tan 89^\circ)$

$\displaystyle \tan \alpha=\cot (90^\circ-\alpha) \rightarrow$ the tangent of an angle is equal to the cotangent of its complementary

So we have $\displaystyle \log (\cot 89^\circ*cot 88^\circ*\ldots*tan 89^\circ)$

Since $\displaystyle \tan \alpha*\cot \alpha = 1$ , we find $\displaystyle \log (\cot 89^\circ*cot 88^\circ*\ldots*tan 89^\circ)=\log (1)=\boxed{0}$

The reader knows that log(a)+log(b)=log(ab). We also know that tan(a)=cot(90-a) and tan(a).cot(a)=1 So, applying this in the problem, we get log(tan1.cot1.tan2.cot2.............tan45), (numbers are in degrees.) this turns out to be log(1) which is equal to "0"

here $\log({\tan{1^{0}}}) + \log({\tan{{89}^{0}}})$ $=\log({\tan{1^{0}}}\times \tan{{89}^{0}})$ $=\log({\frac{\tan({1^{0}})+\tan({89}^{0})}{\tan{{90}^{0}}}}+1)$ $=\log({1})$ $={0}$

And we can do the same thing with second term from beginning with the second term from last and third and so on. Now there are 1+89 = 90 terms and we can make 45 couples and each couple makes value of 0. So, sum up all the values, final answer is $\boxed{0}$

we know $\tan{(A+B)} \space =\space \frac{\tan{(A)}+\tan{(B)}}{1-\tan{(A)}tan{(B)}}$

So, $\tan{(A)}tan{(B)}\space=\space\frac{\tan{(A)}+\tan{(B)}}{\tan{(A+B)} }+1$

log1=0

Since this is the addition of logarithms, an easy way to solve this is to first multiply all of the arguments. This would be log(tan1tan2tan3...tan89). Then, the tangent can be written as sine over cosine. This gives us log(sin1sin2sin3..sin89/cos1cos2cos3..cos89). Since sinx = cos(90-x), the terms can be cancelled out giving us log(1) which equals zero.

Best way to solve it is take log common and all the tan's inside will multiply and the result will be 1 like tan1 tan89=tan1 cot1=1 similarly solve all inside the bracket the we get the fianl value as log1 which is equal to 0