1+Tangents
Geometry
Level
3
Calculate (1+tan1°)×(1+tan2°)×(1+tan3°)×...×(1+tan44°)
The answer is 4194304.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
First If A+B=45° then (1+tanA)(1+tanB)=2
Prove: we all know that
Tan(A+B)=(tanA+tanB)÷(1-tanAtanB) So,tanA+tanB=tan(A+B)-tan(A+B)tanAtanB
(1+tanA)(1+tanB) =1+tanAtanB+tanA+tanB =1+tanAtanB+tan(A+B)-tan(A+B) tanAtanB
Because A+B=45° so tan (A+B)=1 =1+tan(A+B)
Because A+B=45° so tan (A+B)=1 =1+1 =2
So,with the conclusion above,then rearrange the question into a(1+tanA)(1+tanB)form [A+B=45°]
So (1+tan1°)(1+tan44°)(1+tan2°)(1+tan43°)×... ×(1+tan22°)(1+tan23°)=2^22=4194304