Law of Tangent

[nycmathguy]

Why is the Law of Tangent not taught in any trig course?

 

[MathLover1]

The laws of tangent (Law of Tan) describes the relation between difference and sum of sides of a right triangle and tangents of half of the difference and sum of corresponding angles. It represents the relationship between the tangent of two angles of a triangle and the length of the opposite sides.
see the formulas:
https://byjus.com/maths/law-of-tangents/

note, in proof it all comes to sin and cos
since tan(x)=sin(x)/cos(x), it's much easier to use law of sin and cos

 

[nycmathguy]

The laws of tangent (Law of Tan) describes the relation between difference and sum of sides of a right triangle and tangents of half of the difference and sum of corresponding angles. It represents the relationship between the tangent of two angles of a triangle and the length of the opposite sides.
see the formulas:
https://byjus.com/maths/law-of-tangents/

note, in proof it all comes to sin and cos
since tan(x)=sin(x)/cos(x), it's much easier to use law of sin and cos

Interesting. It is easier to avoid the Law of Tangent. Can you work out one example using this law in trigonometry?

 

[MathLover1]

on that link above is one example

for a triangle ABC with sides a, b and c respective to the angles A , B and C is given by,

(a−b)/(a+b)=tan((A−B)/2)/tan((A+B)/2)
Therefore,

(5−3)/(5+3)=tan((A−B)/2)/tan(180°-96°)
2/8=tan((A−B)/2)/tan(84°) ............cross multiply
2*tan(84°)=8tan((A−B)/2)
=>tan(1/2(A−B))=(2/8)tan(42°)=0.2251
=>(1/2)(A−B)=12.7
=>A -B = 25.4°

 

[nycmathguy]

on that link above is one example

for a triangle ABC with sides a, b and c respective to the angles A , B and C is given by,

(a−b)/(a+b)=tan((A−B)/2)/tan((A+B)/2)
Therefore,

(5−3)/(5+3)=tan((A−B)/2)/tan(180°-96°)
2/8=tan((A−B)/2)/tan(84°) ............cross multiply
2*tan(84°)=8tan((A−B)/2)
=>tan(1/2(A−B))=(2/8)tan(42°)=0.2251
=>(1/2)(A−B)=12.7
=>A -B = 25.4°

Very good. Thanks.