Question

If x + y + z = xyz. Then

$\frac{3x-x^3}{1-3x^2}$ + $\frac{3y-y^3}{1-3y^2}$ + $\frac{3z-z^3}{1-3z^3}$ =

• A.
• B. 1
• C. 0
• D. + +

Answer 1

The correct option is A

It is given that x + y + z = xyz

We know in a triangle ABC, tan A + tan B + tan C = tan A .tan B . tan C.

Instead of trying to simplify the given expression, we can use the substitution

x = tan A, y = tan B & z = tan C to solve this in an easy method.

If x = tan A

Then,

$\frac{3x-x^3}{1-3x^2}$ = $\frac{3tanA-ta{n}^{3}A}{1-3ta{n}^{2}A}$ = tan 3A

Similarly, $\frac{3y-y^3}{1-3y^2}$ = $\frac{3tanB-ta{n}^{3}B}{1-3ta{n}^{2}B}$ = tan 3B

and $\frac{3z-z^3}{1-3z^2}$ = $\frac{3tanC-ta{n}^{3}C}{1-3ta{n}^{2}C}$ = tan 3C

So,$\frac{3x-x^3}{1-3x^2}$ + $\frac{3y-y^3}{1-3y^2}$ + $\frac{3z-z^3}{1-3z^2}$ = tan 3A + tan 3B + tan 3C

if A + B + C = 180

3A + 3B + 3C = 3 $\times$ 180= 540

This angle is equivalent to 540-360 =180.

So the three angles 3A, 3B and 3C satisfy the same condition, that the sum of the angles is equal to 180. We can directly say tan 3A + tan 3B + tan 3C = tan 3A . tan 3B . tan 3C

We will try to arrive at this to make sure it was not blunder to write 540 is equivalent to 180.

Taking tan on both sides

3A + 3B + 3C = 3 $\times$ 180= 540

3A + 3B =3$\pi$ - 3C

tan (3A + 3B) = tan (3$\pi$ - 3C) = tan ($\pi$ - 3C)

$\frac{tan3A+tan3B}{1-tan3Atan3B}$ = -tan3C

tan 3A + tan 3B = - tan 3C + tan 3A. tan3B . tan3C

tan 3A + tan 3B + tan 3C = tan 3A . tan 3B . tan3C

So, $\frac{3tanA-ta{n}^{3}A}{1-3ta{n}^{2}A}$ + $\frac{3tanB-ta{n}^{3}B}{1-3ta{n}^{2}B}$ + $\frac{3tanC+ta{n}^{3}C}{1-3ta{n}^{2}C}$

= $\frac{3tanA-ta{n}^{3}A}{1-3ta{n}^{2}A}$ $\times$ $\frac{3tanB-ta{n}^{3}B}{1-3ta{n}^{2}B}$ $\times$ $\frac{3tanC-ta{n}^{3}C}{1-3ta{n}^{2}C}$

Replace tan A = x, tan B = y, tanC = z

$\frac{3x-x^3}{1-3x^2}$ + $\frac{3y-y^3}{1-3y^2}$ + $\frac{3z-z^3}{1-3z^3}$ = $\frac{3x-x^3}{1-3x^2}$ $\times$ $\frac{3y-y^3}{1-3y^2}$ $\times$ $\frac{3z-z^3}{1-3z^2}$