Question

tanA/2=√(s-b)(s-c)/s(s-a) prove that.​

Answer 1

Answer:

the hero's formula

we have ,

2 then first of all we write

s=2/2=0

then the :

hero's formula you appiled

hope its helpful to you

Answer 2

Answer:

half-angle formula.

If ABC is a triangle, A, B, and C are the three angles of the triangle and a,b,c are the sides opposite to the corresponding angles and s is the semi-perimeter.

s = (a + b + c)/2 then tan A/2 = (s−b)(s−c)/s(s−a)

Proof :

First, let’s prove the half-angle formula for cos A/2

Using the cosine law :

2bc cos A = b² + c² - a²

=> 2bc + 2bc cos A = 2bc + b² + c² - a²

=> 2bc (1 + cos A) = (b + c)² - a²

Now using the trigonometric sub-multiple angle formula :

2bc * 2cos² A/2 = (b +c + a)(b +c - a)

=> 4bc cos² A/2 = 2s * (2s - 2a) (because : a + b + c = 2s)

=> cos A/2 = s(s−a)/bc

Now, let us prove the half-angle formula for sin A/2 using the cosine law :

- 2bc cos A = a² - ( b² + c²)

=> 2bc - 2bc cos A = 2bc + a² - ( b² + c²)

=> 2bc(1 - cos A) = a² - (b² - c²)

=> 2bc * 2sin² A/2 = (a - b + c)(a + b - c) = (2s - 2b)(2s - 2c)

=> sin A/2 = (s−b)(s−c)/bc

We have calculated sin A/2 and cos A/2

We know that tan A/2 = sin A/2 / cos A/2

tan A/2 = (s−b)(s−c)/bc √ / s(s−a)/bc

Hence, tan A/2 = (s−b)(s−c)/s(s−a)

√ (because : bc will be canceled out)