Given that A, B, C, are three ∠’s of a ∆ therefore
A + B + C = π
Also A = π/4 ⇒ B + C = 3 π/4
⇒ 0 < B, C < 3π/4
Now tan B tan C = P
⇒ sin B sin C/cos B cos C = p/1
Applying componendo and dividendo, we get
Sin B sin C + cos B cos C/cos B cos C – sin B sin C = 1+p/1-p
⇒ cos (B – C)/cos (B + C) = 1 + p/1 – p
⇒ cos (B – C) = 1 + p/1 – p (-1/√2) . . . . . . . . . . . . . (1) [∵ B + C = 3 π/4]
Now, as B and C can vary from 0 to 3π/4
∴ 0 ≤ B – C < 3π/4
⇒ 1/√2 < cos (B – C) ≤ 1
From eq” (1) substituting the value of cos (B – C), we get
-1/√2 < 1 + p/√2(p – 1) ≥ 1
⇒ -1/√2 < 1 + p/√2(p – 1) and 1 + p/√2(p – 1) ≤1
⇒ 0 < 1 + p + 1/p – 1 and (p + 1) - √2 (p – 1)/√2 (p – 1) ≤0
⇒ 2p/p – 1 > 0 and p + 1 - √2 p + √2 / √2(p – 1) ≤ 0
⇒ p (p – 1) > 0 and (1 - √2) p + (√2 + 1)/(p – 1) ≤ 0
⇒ p∈ ( -∞, 0) ∪ (1, ∞), and –p + (√2 + 1)2/(p – 1) ≤ 0
⇒ [p – (3 + 2√2)] [p – 1] ≥ 0
Combining the two cases, we get
p∈ (- ∞, 0) ∪ [3 + 2 √2, ∞).
