SOLUTION: CotA =4/3 and (a+b)=90degree find the value of tanB = ?

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Question 1042126: CotA =4/3 and (a+b)=90degree find the value of tanB = ?
Found 2 solutions by Aldorozos, ikleyn:
Answer by Aldorozos(172)   (Show Source): You can put this solution on YOUR website!
Tan(A+B) = (TanA+TanB)/(1-TanATanB)
Please Watch this video:
https://www.youtube.com/watch?v=igNTc-G_77k
We know that Tan = 1/cot
If (a+b) = 90 then tan(a+b)= tan 90
We know that tan 90 = 0 then tan(a+b)= 0
We know that Tan(A+B) = (TanA+TanB)/(1-TanATanB)= 0 for
(TanA+TanB)/(1-TanATanB)to be equal to zero the numerator has to be equal to zero which means TanA+TanB = 0. We already know what the tan of a is (Tan a = 1/CotA) This means that tan a = 1/(4/3) = 3/4 (Please note that the problem has already given us the value of Cot A which is 4/3
Therefore Tan A + Tan B = 0 replacing Tan A with 3/4 we get
Tan B + 3/4 = 0. Therefore Tan B = -3/4

Answer by ikleyn(52908)   (Show Source): You can put this solution on YOUR website!
.
Cot(A) = 4/3 and (A+B) = 90 degree. Find the value of tan(B) = ?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

Since A + B = 90°, you have B = 90° - A.

Now, tan(B) = tan(90°-A) = cot(A).

It is a standard trigonometry identity (see, for example, this site). 

Therefore, tan(B) = cot(A) = .

The solution of the other tutor contains very serious error, so disregard it.

In particular, he writes

    "If (a+b) = 90 then tan(a+b)= tan 90
    We know that tan 90 = 0 then tan(a+b)= 0"

which is totally wrong. Everybody knows that tan(90°) is not defined ("is infinity").

This is why I wrote this post for you.


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