If $x + y + z = xyz$, then prove that $\dfrac{2x}{1 - x^2} + \dfrac{2y}{1 - y^2} + \dfrac{2z}{1 - z^2} = \dfrac{2x}{1 - x^2}\cdot\dfrac{2y}{1 - y^2}\cdot\dfrac{2z}{1 - z^2}$.

Question

Accepted Answer

> [!TLDR]
>
> This is the tangent double-angle formula in disguise: with $x=	an\alpha$ etc., $\frac{2x}{1-x^2}=	an2\alpha$, and the given condition $x+y+z=xyz$ forces $\alpha+\beta+\gamma$ to be a multiple of $\pi$, which is exactly the condition under which $	an2\alpha+	an2\beta+	an2\gamma=	an2\alpha	an2\beta	an2\gamma$.

**Step 1. A general fact about angles summing to a multiple of $\pi$.** If $P+Q+R=m\pi$ for an integer $m$, then $R=m\pi-(P+Q)$, so $	an R=	an\big(m\pi-(P+Q)\big)=-	an(P+Q)=-\dfrac{	an P+	an Q}{1-	an P	an Q}$. Cross-multiplying: $	an R(1-	an P	an Q)=-(	an P+	an Q)$, i.e. $	an R-	an P	an Q	an R=-	an P-	an Q$, which rearranges to $$	an P+	an Q+	an R=	an P	an Q	an R.$$

**Step 2. Substitute $x=	an\alpha,\ y=	an\beta,\ z=	an\gamma$.** The hypothesis $x+y+z=xyz$ becomes $	an\alpha+	an\beta+	an\gamma=	an\alpha	an\beta	an\gamma$, which (reading Step 1 in reverse) is exactly the condition $\alpha+\beta+\gamma=n\pi$ for some integer $n$ (since $	an(\alpha+\beta+\gamma)=0$ precisely when $\alpha+\beta+\gamma$ is a multiple of $\pi$, and the given equation is the numerator of the triple-angle tangent-sum expansion set to zero).

**Step 3. Double the angle condition.** From $\alpha+\beta+\gamma=n\pi$, doubling gives $2\alpha+2\beta+2\gamma=2n\pi$, which is again an integer multiple of $\pi$ (with $m=2n$). …

Q.If $x + y + z = xyz$ , then prove that $\dfrac{2x}{1 - x^2} + \dfrac{2y}{1 - y^2} + \dfrac{2z}{1 - z^2} = \dfrac{2x}{1 - x^2}\cdot\dfrac{2y}{1 - y^2}\cdot\dfrac{2z}{1 - z^2}$ .

Concept understanding — Sum-to-Product & Product-to-Sum

🎓Unlock everything free for 14 days

Unlock everything free for 14 days