In triangle ABC Prove that

Question

in triangle ABC prove that: tanA/tanB = a²+c²-b² /b²+c²-a²

Answer 1

Using trigonometric identities---

 

a^2+c^2-b^2= 2*a*c*cos(B)

 

Likewise---

 

b^2+c^2-a^2=2*b*c*cos(A)

 

So, the right hand side given expression =2accos(B)/2bccos(A)

 

                                                           =acos(B)/bcos(A)

 

We know----

Sin(A)/a. =. Sin(B)/b

 

Using the above relationship---

 

a. = bsin(A)/sin(B)

 

the right hand expression = b sin(A)cos(B)/bsin(B)cos(A)

 

                                      = tan(A)/tan(B)

              

                                      = left hand side

 

Any questions,welcome

Answer 2

How about this way?

 

 

Let's say tan A =  (sin A) / (cos A) and tan B = (sin B) / (cos B).

And even further back, tan A = a / b and tan B = b / a.

 

 

From identities sin (90 - x) = cos x and cos (90 - x) = sin x.

So, you put in as sin (90 - A) = cos A, which is sin B.  sin B = cos A & sin A = cos B.

 

 

From Law of Sines: (sin A) / a = (sin B) / b

From Law of Cosines: 

 

a^{2 =} b²+c² - 2bc cos A

b^{2 =} a2+c^{2 -} 2ac cos B

 

 

Rewrite as in terms of cos A & cos A.

(a²+c²-b²) / (2ac) = cos B & (b²+c²-a²) / (2bc) = cos A.

Use the identities above: sin (90 - x) = cos x and cos (90 - x) = sin x. 

 

Which means sin a = cos b & sin b = cos a.

Here are the calculating and evaluating parts come in:

As you divide (a2+c2-b2) / (2ac) = cos B & (b²+c²-a²) / (2bc) = cos A, it will give you
(a2+c2-b2) b / a (b2+c2-a2). Rewrite it as [(a²+c²-b²) / (b²+c²-a²)] * (b / a).
And if you evaluate (cos B) / (cos A), you will get (cot B) * (b / a) or (1/tan B) * (b / a).
You can catch that (1/tan B) * (b / a). So, (cos B) / (cos A) = (1/tan B) * (b / a).

 

 

Revisit the original problem: tan A / tan B
To make tan A / tan B, both side needs to be multiplied by either tan A or (a / b). As it multiplies, (b/a) * (a/b) cancels each other out and leaves it with (a / b) * (cos B) / (cos A) = [(a²+c²-b²) / (b²+c²-a²)].

 

 

Rewrite this as: (a / b) * (1 / tan B) = [(a²+c²-b²) / (b²+c²-a²)].
Rewrite again as: (tan A) * (1 / tan B) = [(a²+c²-b²) / (b²+c²-a²)].
Which means tan A / tan B = [(a²+c²-b²) / (b²+c²-a²)].

 

 

I know it is long but you will see all the details. All the work is done from left side to verify the right side.