Question

A+B+C=180 degree then tan^2 A\2+tan^2B/2+tanC/2=?

Answer 1

Answer:

Step-by-step explanation:

A+B+C = pi

so A/2 + B/2 + C/2 = pi/2

or A/2 + B/2 = (pi/2- C/2)

take tan of both sides

tan (A/2 + B/2) = tan (pi/2-C/2) = cot C/2 = 1/ tan C/2

or (tan A/2 + tan B/2)/(1- Tan A/2 tan B/2) = 1/tan C/2

or tan C/2 tan A/2 + tan C/2 tan B/2 = 1 - tan A/2 tan B/2

or tan A/2 tan B/2 + tan B/2 tan C/2 + tan A/2 tan C/2 = 1

proved

Hope this answers your question

Please mark as brainliest :)

Answer 2

Given :  A+B+C=180°

To find :  Prove that    ( Tan²(A/2) + Tan²(B/2) + Tan²(c/2) ≥ 1

Solution:

( Tan (A/2) - Tan(B/2))² + ( Tan (B/2) - Tan(C/2))² +  (Tan (C/2) - Tan(A/2))² ≥ 0

=> 2(Tan²(A/2) + Tan²(B/2) + Tan²(C/2) ) - 2(Tan(A/2)Tan(B/2) + Tan(B/2)Tan(c/2) + Tan(C/2)Tan(A/2)  ≥ 0

=> 2(Tan²(A/2) + Tan²(B/2) + Tan²(C/2) )  ≥  2(Tan(A/2)Tan(B/2) + Tan(B/2)Tan(c/2) + Tan(C/2)Tan(A/2)

=> (Tan²(A/2) + Tan²(B/2) + Tan²(C/2) )  ≥  (Tan(A/2)Tan(B/2) + Tan(B/2)Tan(c/2) + Tan(C/2)Tan(A/2)      Eq1

A + B + C = 180°

=> A/2 + B/2 + C/2 =90°

=> A/2 + B/2 = (90°- C/2)

Tan (A/2 + B/2) =Tan(90°--C/2)

Tan(90°--C/2)  = cot C/2 = 1/ tan C/2

=>  (tan A/2 + tan B/2)/(1- Tan A/2 tan B/2) = 1/tan C/2

=> tan C/2 tan A/2 + tan C/2 tan B/2 = 1 - tan A/2 tan B/2

=>  tan A/2 tan B/2 + tan B/2 tan C/2 + tan A/2 tan C/2 = 1

using this in eq1

=>  (Tan²(A/2) + Tan²(B/2) + Tan²(C/2) )  ≥ 1

QED

Hence proved

Learn more:

If sec theta + tan theta = p, show that sec theta - Brainly.in

https://brainly.in/question/2142440

in triangle abc prove that B minus C cot a by 2 + c minus A cot B by 2 ...

https://brainly.in/question/12878751