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[SOLVED] Proving two sides of equation for triangles
In angle ABC, which is an isosceles triangle with <B = <C, show that
2cot(a) = tan(b) = cot(b)
tan2a = 2tana / 1 - tan^2 a
tan (x - y) = tanx - tany / 1 + tanx tany
Since it is isosceles, that means two sides are equal, therefore, <A = 180 - 2B
2cot(A) = 2cot (180 - 2B)
= 1 / 2tan(180 - 2B)
= 1 / 2(tan180 - tan2B / 1 + tan180 tan2B)
= 1 / 2(-tan2B)
= 1 / -2tan2B)
= 1 / -2(2tanB / 1 - tan^2 B)
After this, I get confused. Can someone please tell me if I am doing this right? Please help. Thanks.
Homework Statement
In angle ABC, which is an isosceles triangle with <B = <C, show that
2cot(a) = tan(b) = cot(b)
Homework Equations
tan2a = 2tana / 1 - tan^2 a
tan (x - y) = tanx - tany / 1 + tanx tany
The Attempt at a Solution
Since it is isosceles, that means two sides are equal, therefore, <A = 180 - 2B
2cot(A) = 2cot (180 - 2B)
= 1 / 2tan(180 - 2B)
= 1 / 2(tan180 - tan2B / 1 + tan180 tan2B)
= 1 / 2(-tan2B)
= 1 / -2tan2B)
= 1 / -2(2tanB / 1 - tan^2 B)
After this, I get confused. Can someone please tell me if I am doing this right? Please help. Thanks.