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

Explanation

`A+B+C=pi`
`rArr B+C=(3pi)/(4)rArr0ltB,Clt(3pi)/(4)`
Also `tan B tan C=p`
or `(sin B sin C)/(cos B cos C)=(p)/(1)`
or `(cos B cos C-sin B sinC)/(cos B cos C+sin B sinC)=(1-p)/(1+p)`
or `(cos(B+C))/(cos(B-C))=(1-p)/(1+p)`
or `(1+p)/(sqrt(2)(P-1))=cos(B-C)`
Since B or C can very from 0 to `3pi//4`, we get
`-(3pi)/(4)ltB-Clt(3pi)/(4)` or `-(1)/sqrt(2)ltcos(B-C)le1`.
Equation i will now lead to `-(1)/sqrt(2)lt(p+1)/(sqrt(2)(p-1))le1`
For `0lt1+(p+1)/(p-1)`
or `(2p)/((p-1))gt0`
`therefore plt0` or `pgt1`
Also `(p+1-sqrt(2)(p-1))/(sqrt(2)(p-1))`
or `((p-(sqrt(2)+1)^(2)))/((P-1))ge0`
`therefore plt1` or `pge(sqrt(2)+1)^(2)`
Combining Eqs. (ii) and (iii), we get `plt0` or `pge(sqrt(2)+1)^(2)`.