Show that if A, B, and C are the angles of an acute triangle, then tan A + tan B + tan C = tan A tan B tan C.
I tried drawing perpendiculars and stuff but it doesn't seem to work?
For me, the trig identities don't seem to plug in as well.
Help is appreciated, thanks.
Thanks a ton!
C = 180-(A+B), so
tan C = -tan(A+B) = (tanA+tanB)/(tanAtanB-1)
tanA+tanB+tanC
= tanA+tanB + (tanA+tanB)/(tanAtanB-1)
= (tanA+tanB)(1 + 1/(tanAtanB-1))
= (tanA+tanB)(tanAtanB)/(tanAtanB-1)
= (tanAtanB)(tanA+tanB)/(tanAtanB-1)
= tanAtanBtanC
To prove the given statement, we will make use of the basic trigonometric identity:
tan(A + B) = (tan A + tan B) / (1 - tan A tan B).
First, let's write the expression we want to prove:
tan A + tan B + tan C = tan A tan B tan C.
We know that the angles A, B, and C are the angles of an acute triangle, which means that they are all less than 90 degrees. Furthermore, we also know that the sum of the angles in a triangle is 180 degrees:
A + B + C = 180.
Now, consider the sum of two angles A and (B + C):
A + (B + C) = 180.
We can rewrite this equation as:
A = 180 - (B + C).
Let's substitute this expression for A into our original equation:
tan(180 - (B + C)) + tan B + tan C = tan(180 - (B + C)) tan B tan C.
Next, we will use the trigonometric identity mentioned earlier:
tan(180 - (B + C)) = (tan 180 - tan (B + C)) / (1 + tan 180 tan (B + C)).
Since tan 180 = 0, we can simplify the above expression:
tan(180 - (B + C)) = - tan (B + C).
Substituting this back into our equation:
- tan (B + C) + tan B + tan C = - tan (B + C) tan B tan C.
Multiplying both sides of the equation by -1 yields:
tan (B + C) - tan B - tan C = tan (B + C) tan B tan C.
Now, let's rearrange this equation:
tan (B + C) + (-tan B - tan C) = tan (B + C) tan B tan C.
Using the fact that tan (-x) = -tan x:
tan (B + C) - tan(-B) - tan(-C) = tan (B + C) tan B tan C.
Finally, applying the trigonometric identity again:
(tan (B + C) - tan(-B))(1 - tan(-C)tan(B + C)) = tan (B + C) tan B tan C.
Simplifying the expression:
tan (B + C) - tan B (1 + tan B tan C) = tan (B + C) tan B tan C.
Further simplifying:
tan (B + C) - tan B - tan B tan C = tan (B + C) tan B tan C.
Now, we can cancel out tan (B + C) from both sides:
- tan B - tan B tan C = - tan B tan C.
Finally, dividing both sides by -1:
tan B + tan B tan C = tan B tan C.
This proves that if A, B, and C are the angles of an acute triangle, then tan A + tan B + tan C = tan A tan B tan C.