Logs and Trig?

By the property of logarithms $\ln a+\ln b=\ln(ab)$ the expression is now: $\ln(\tan 1° \tan 2° \tan 3° ... \tan 89°)$ . Now, by the identity $\tan(90°-\theta)=\dfrac{1}{\tan \theta}$ the expression is now: $\ln\left(\tan 1° \tan 2° \tan 3° ... \tan 45° \dfrac{1}{\tan 1°} \dfrac{1}{\tan 2°} \dfrac{1}{\tan 3°} ...\right)$ . Everything cancels except $\tan 45°$ which is $1$ , so the expression is now $\ln 1=\boxed{0}$ .

As we know $\ln a+\ln b=\ln a\times b$ so the series will go on as $\ln (\tan 1\times tan2\times tan3 \times \:....\:\times tan89)$ Now from the multiplication we get the result, $\ln 1=\boxed{0}$

The Question actually involves both trigonometry and logarithmic knowledge. Clearly ln(x) + ln(y) = ln(xy), therefore, ln(tan1.tan2.....tan89)=ln(tan1tan2.....tan45.1/tan1.tan2.....)=ln(tan45)=ln1=0.

ln(tan 1 ) + ln(tan 89) = 0 same as ln(tan2) + ln(tan 88) = 0 sum up all the values = 0

So,

In (tan 1 * tan 2 * tan 3 * ... * tan 89) in degrees all is, 1..

Then,

In(1) = 0

Final answer: 0