Question

If in a triangle ABC, we define \[x=\tan \left( \dfrac{B-C}{2} \right)\tan \left( \dfrac{A}{2} \right)\], $y=\tan \left( \dfrac{C-A}{2} \right)\tan \left( \dfrac{B}{2} \right)$, $z=\tan \left( \dfrac{A-B}{2} \right)\tan \left( \dfrac{C}{2} \right)$, then show that x + y + z = – xyz.

Answer 1

Hint: We start solving the problem by using half angle formula from the properties of triangle. We use these formulas to find the values of x, y and z. We convert all these x, y and z in the form of $\dfrac{1+x}{1-x},\dfrac{1+y}{1-y},\dfrac{1+z}{1-z}$. We multiply $\dfrac{1+x}{1-x},\dfrac{1+y}{1-y},\dfrac{1+z}{1-z}$ to each other and make necessary calculations to get the required result.

Complete step by step answer:
Given that we have a triangle ABC and we define \[x=\tan \left( \dfrac{B-C}{2} \right)\tan \left( \dfrac{A}{2} \right)\], $y=\tan \left( \dfrac{C-A}{2} \right)\tan \left( \dfrac{B}{2} \right)$, $z=\tan \left( \dfrac{A-B}{2} \right)\tan \left( \dfrac{C}{2} \right)$. We need to prove that x + y + z = – xyz.
Let us draw the triangle ABC,

We know that from properties of triangles we have $\tan \left( \dfrac{B-C}{2} \right)=\dfrac{\left( b-c \right)}{\left( b+c \right)}.\cot \left( \dfrac{A}{2} \right)$ ---(1).
According to the problem, \[x=\tan \left( \dfrac{B-C}{2} \right)\tan \left( \dfrac{A}{2} \right)\].
From equation (1),
$\Rightarrow $ \[x=\dfrac{\left( b-c \right)}{\left( b+c \right)}.\cot \left( \dfrac{A}{2} \right).\tan \left( \dfrac{A}{2} \right)\].
$\Rightarrow $ $x=\dfrac{\left( b-c \right)}{\left( b+c \right)}$.
$\Rightarrow $ $\left( 1+x \right)=1+\left( \dfrac{b-c}{b+c} \right)$.
$\Rightarrow $ $\left( 1+x \right)=\dfrac{\left( b+c \right)+\left( b-c \right)}{\left( b+c \right)}$.
$\Rightarrow $ $\left( 1+x \right)=\dfrac{2b}{\left( b+c \right)}$ ---(2).
$\Rightarrow $ $\left( 1-x \right)=1-\left( \dfrac{b-c}{b+c} \right)$.
$\Rightarrow $ $\left( 1-x \right)=\dfrac{\left( b+c \right)-\left( b-c \right)}{\left( b+c \right)}$.
$\Rightarrow $ $\left( 1-x \right)=\dfrac{2c}{\left( b+c \right)}$ ---(3).
Using equations (2) and (3), $\dfrac{\left( 1+x \right)}{\left( 1-x \right)}=\dfrac{\left( \dfrac{2b}{b+c} \right)}{\left( \dfrac{2c}{b+c} \right)}$.
$\Rightarrow $ $\dfrac{\left( 1+x \right)}{\left( 1-x \right)}=\dfrac{2b}{2c}$ ---(4).
We know that from properties of triangles we have $\tan \left( \dfrac{C-A}{2} \right)=\dfrac{\left( c-a \right)}{\left( c+a \right)}.\cot \left( \dfrac{B}{2} \right)$ ---(5).
According to the problem, \[y=\tan \left( \dfrac{C-A}{2} \right)\tan \left( \dfrac{B}{2} \right)\].
From equation (5),
$\Rightarrow $ \[y=\dfrac{\left( c-a \right)}{\left( c+a \right)}.\cot \left( \dfrac{B}{2} \right).\tan \left( \dfrac{B}{2} \right)\].
$\Rightarrow $ $y=\dfrac{\left( c-a \right)}{\left( c+a \right)}$.
$\Rightarrow $ $\left( 1+y \right)=1+\left( \dfrac{c-a}{c+a} \right)$.
$\Rightarrow $ $\left( 1+y \right)=\dfrac{\left( c+a \right)+\left( c-a \right)}{\left( c+a \right)}$.
$\Rightarrow $ $\left( 1+y \right)=\dfrac{2c}{\left( c+a \right)}$ ---(6).
$\Rightarrow $ $\left( 1-y \right)=1-\left( \dfrac{c-a}{c+a} \right)$.
$\Rightarrow $ $\left( 1-y \right)=\dfrac{\left( c+a \right)-\left( c-a \right)}{\left( c+a \right)}$.
$\Rightarrow $ $\left( 1-y \right)=\dfrac{2a}{\left( c+a \right)}$ ---(7).
Using equations (6) and (7), $\dfrac{\left( 1+y \right)}{\left( 1-y \right)}=\dfrac{\left( \dfrac{2c}{c+a} \right)}{\left( \dfrac{2a}{c+a} \right)}$.
$\Rightarrow $$\dfrac{\left( 1+y \right)}{\left( 1-y \right)}=\dfrac{2c}{2a}$ ---(8).
We know that from properties of triangles we have $\tan \left( \dfrac{A-B}{2} \right)=\dfrac{\left( a-b \right)}{\left( a+b \right)}.\cot \left( \dfrac{C}{2} \right)$ ---(9).
According to the problem, \[z=\tan \left( \dfrac{A-B}{2} \right)\tan \left( \dfrac{C}{2} \right)\].
From equation (9),
$\Rightarrow $ \[z=\dfrac{\left( a-b \right)}{\left( a+b \right)}.\cot \left( \dfrac{C}{2} \right).\tan \left( \dfrac{C}{2} \right)\].
$\Rightarrow $ $z=\dfrac{\left( a-b \right)}{\left( a+b \right)}$.
$\Rightarrow $ $\left( 1+z \right)=1+\left( \dfrac{a-b}{a+b} \right)$.
$\Rightarrow $ $\left( 1+z \right)=\dfrac{\left( a+b \right)+\left( a-b \right)}{\left( a+b \right)}$.
$\Rightarrow $ $\left( 1+z \right)=\dfrac{2a}{\left( a+b \right)}$ ---(10).
$\Rightarrow $ $\left( 1-z \right)=1-\left( \dfrac{a-b}{a+b} \right)$.
$\Rightarrow $ $\left( 1-z \right)=\dfrac{\left( a+b \right)-\left( a-b \right)}{\left( a+b \right)}$.
$\Rightarrow $ $\left( 1-z \right)=\dfrac{2b}{\left( a+b \right)}$ ---(11).
Using equations (10) and (11), $\dfrac{\left( 1+z \right)}{\left( 1-z \right)}=\dfrac{\left( \dfrac{2a}{a+b} \right)}{\left( \dfrac{2b}{a+b} \right)}$.
$\Rightarrow $ $\dfrac{\left( 1+z \right)}{\left( 1-z \right)}=\dfrac{2a}{2b}$ ---(12).
We multiply all the results obtained from equations (4), (8) and (12), $\dfrac{\left( 1+x \right)}{\left( 1-x \right)}\times \dfrac{\left( 1+y \right)}{\left( 1-y \right)}\times \dfrac{\left( 1+z \right)}{\left( 1-z \right)}=\left( \dfrac{2b}{2c} \right)\times \left( \dfrac{2c}{2a} \right)\times \left( \dfrac{2a}{2b} \right)$.
$\Rightarrow $ $\dfrac{\left( 1+x \right)\times \left( 1+y \right)\times \left( 1+z \right)}{\left( 1-x \right)\times \left( 1-y \right)\times \left( 1-z \right)}=\dfrac{8abc}{8abc}$.
$\Rightarrow $ $\dfrac{\left( 1+x+y+xy \right)\times \left( 1+z \right)}{\left( 1-x-y+xy \right)\times \left( 1-z \right)}=1$.
$\Rightarrow $ $\dfrac{\left( 1+x+y+xy+z+xz+yz+xyz \right)}{\left( 1-x-y+xy-z+xz+yz-xyz \right)}=1$.
$\Rightarrow $ $1+x+y+z+xy+xz+yz+xyz=1-x-y-z+xy+xz+yz-xyz$.
$\Rightarrow $ $1-1+x+y+z+x+y+z+xy+yz+xz-xy-yz-zx=-xyz-xyz$.
$\Rightarrow $ \[2x+2y+2z=-2xyz\].
$\Rightarrow $ \[x+y+z=-xyz\].
∴ We have proved \[x+y+z=-xyz\].

Note: We should not solve for x, y and z individually in the middle of the problem again, as it is already given in the problem. Similarly, we get problems to find the value of A, B and C if the given condition is satisfied.