Question

in triangle ABC. p.t.
i)Cos (A+B) + Cosc =0.
ii) Cos (A+B/2) = Sin(C/2)
iii) Tan (A+B/2) = Cot (C/2)​

Answer 1

Answer:

Step-by-step explanation:

(I) A+B+C = 180

   A+B = 180-C

Taking Cos on both the sides

Cos(A+B) = Cos(180 - C)

Cos(A+B) = -Cos C  {Cos (180-x)  = -Cos x }

Therefor putting the value of Cos(A+B)

-Cos C + Cos C = 0 [Hence proved]

(II) A+B+C = 180

    A+B = 180-C

    A+B/2 = 90 - c/2

Taking Cos on both the sides

Cos(A+B/2) = Cos(90- c/2)

Cos(A+B/2) = Sin(c/2)   { Cos(90-x) = Sin x }

(III)A+B+C = 180

    A+B = 180-C

    A+B/2 = 90 - c/2

Taking Tan on both the sides

Tan(A+B/2) = Tan(90- c/2)

Tan(A+B/2) = Cot(c/2)   { Tan(90-x) = Cot x }