# Let A, B, C be three angles such that A = π/4  and tanB tanC = p. Find all possible values of p …

Q: Let A, B, C be three angles such that A = π/4  and tanB tanC = p. Find all possible values of p such that A, B, C are the angles of a triangle.

Solution : A + B + C = π

B + C = π – A

B + C = π – π/4 = 3π/4

0 < B , C < 3π/4

tanB tanC = p

$\large \frac{sinB sinC}{cosB cosC} = \frac{p}{1}$

$\large \frac{cosB cosC – sinB sinC}{cosB cosC + sinB sinC } = \frac{1-p}{1+p}$

$\large \frac{cos(B+C)}{cos(B-C)} = \frac{1-p}{1 + p}$

$\large \frac{cos(3\pi/4)}{cos(B-C)} = \frac{1-p}{1 + p}$

$\large cos(B-C) = \frac{1+p}{\sqrt{2}(p-1)}$ …(1)

Since B or C can vary from 0 to 3π/4

$\large -\frac{1}{\sqrt{2}} < cos(B-C) \le 1$

Equation (1) will now lead to

$\large -\frac{1}{\sqrt{2}} < \frac{1+p}{\sqrt{2}(p-1)} \le 1$ $1 + \frac{1+p}{(p-1)} > 0$

$\large \frac{2p}{p-1} > 0$

p < 0 , p > 1 …(2)

Also , $\large \frac{p+1 – \sqrt{2}(p-1)}{\sqrt{2}(p-1)} \le 0$

$\large \frac{(p-(\sqrt{2}+1)^2)}{p-1} \ge 0$

p < 1 or $p \ge (\sqrt{2}+1)^2$ …(3)

Combining (2) & (3) , we get

p < 0 or , $p \ge (\sqrt{2}+1)^2$