Question

If B+C < π/2, then

a) tanBtanC < 1
b) tanBtanC > 1​

Answer 1

Answer:

Step-by-step explanation:

3 tanФ = Cot Ф = 1/tanФ

3 tan²Ф = 1

TanФ = + 1/√3  

Ф = n π + π/6

Ф = π/6 or π+π/6 = 7π/6 or π-π/6 = 5π/6 or 2π-π/6 = 11π/6

There are 4 solutions for 0° < Ф < 360°.

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A + B + C = π/2

 => Tan(A+B) = Tan(π/2-C) = Cot C = 1/TanC

 => Tan (B+C) = 1/TanA

 => Tan(C+A) = 1/TanB

Tan (A+B) = [TanA + TanB]/[1 - TanA TanB]

Rearrange:  TanA TanB = 1 - (TanA+TanB)/Tan(A+B)

                                     = 1 - (TanA+TAnB) ×TanC     ---(1)

Similarly:   Tan B TanC = 1 - (TanB + TanC) × TanA  ---(2)

                 Tan C TanA = 1 - (TanC + TanA) ×TanB  --- (3)

Add (1) , (2) & (3):

LHS = TanA TanB + TanB TanC + TanC Tan A

       = 3 - 2 (TanA TanB + TanB TanC + Tanc TanA)

       = 3 -  2 LHS

So  TanA TanB + TanB TanC + Tan C TanA = 1

Answer 2

Answer:

Hlo

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