Question

If $\alpha +\beta +\gamma ={180}^{\circ }$and $cot\theta =cot\alpha +cot\beta +cot\gamma ,0<\theta <{180}^{\circ }$

Show that${\sin }^3\theta =\sin (\alpha -\theta )\sin (\beta -\theta )\sin (\gamma -\theta )$

Short answer

If $\alpha +\beta +\gamma ={180}^{\circ }$and $cot\theta =cot\alpha +cot\beta +cot\gamma ,0<\theta <{180}^{\circ }$

Then

$$\begin{gathered} \sin (\alpha -\theta )=\frac{{\sin }^2\alpha \sin \theta }{\sin \beta \sin \gamma } \\ \sin (\beta -\theta )=\frac{{\sin }^2\beta \sin \theta }{\sin \alpha \sin \gamma } \\ \sin (\gamma -\theta )=\frac{{\sin }^2\gamma \sin \theta }{\sin \beta \sin \alpha } \end{gathered}$$

On multiplying all equations

$$\begin{gathered} \sin (\alpha -\theta )\sin (\beta -\theta )\sin (\gamma -\theta )=\frac{{\sin }^2\alpha \sin \theta }{\sin \beta \sin \gamma }\times \frac{{\sin }^2\beta \sin \theta }{\sin \alpha \sin \gamma }\times \frac{{\sin }^2\gamma \sin \theta }{\sin \beta \sin \alpha } \\ \sin (\alpha -\theta )\sin (\beta -\theta )\sin (\gamma -\theta )={\sin }^2\theta \end{gathered}$$

Step-by-step solution

Step 1. Given data

The given equations are

$$\begin{gathered} \alpha +\beta +\gamma ={180}^{\circ } \\ cot\theta =cot\alpha +cot\beta +cot\gamma \\ 0<\theta <{180}^{\circ } \end{gathered}$$

The equation we need to prove is

${\sin }^3\theta =\sin (\alpha -\theta )\sin (\beta -\theta )\sin (\gamma -\theta )$

Step 2. Formation of the equation for proof

As $cot\theta =cot\alpha +cot\beta +cot\gamma$

So

$$\begin{gathered} cot\theta =cot\alpha +cot\beta +cot\gamma \\ cot\theta -cot\alpha =cot\beta +cot\gamma \\ \frac{\cos \theta }{\sin \theta }-\frac{\cos \alpha }{\sin \alpha }=\frac{\cos \beta }{\sin \beta }+\frac{\cos \gamma }{\sin \gamma } \\ \frac{\cos \theta \sin \alpha -\cos \alpha \sin \theta }{\sin \theta \sin \alpha }=\frac{\cos \beta \sin \gamma +\cos \gamma \sin \beta }{\sin \beta \sin \gamma } \\ \frac{\sin (\alpha -\theta )}{\sin \theta \sin \alpha }=\frac{\sin (\gamma +\beta )}{\sin \beta \sin \gamma } \\ and\alpha +\beta +\gamma ={180}^{\circ } \\ \beta +\gamma ={180}^{\circ }-\alpha \\ So\frac{\sin (\alpha -\theta )}{\sin \theta \sin \alpha }=\frac{\sin ({180}^{\circ }-\alpha )}{\sin \beta \sin \gamma } \\ \frac{\sin (\alpha -\theta )}{\sin \theta \sin \alpha }=\frac{\sin \left(\alpha \right)}{\sin \beta \sin \gamma } \\ \sin (\alpha -\theta )=\frac{{\sin }^2\alpha \sin \theta }{\sin \beta \sin \gamma }\left(i\right) \end{gathered}$$

Similarly

$$\begin{gathered} \sin (\beta -\theta )=\frac{{\sin }^2\beta \sin \theta }{\sin \alpha \sin \gamma }\left(ii\right) \\ \sin (\gamma -\theta )=\frac{{\sin }^2\gamma \sin \theta }{\sin \beta \sin \alpha }\left(iii\right) \end{gathered}$$

Step 3. proof

Multiply equation(i),(ii),and(iii)

$$\begin{gathered} \sin (\alpha -\theta )\sin (\beta -\theta )\sin (\gamma -\theta )=\frac{{\sin }^2\alpha \sin \theta }{\sin \beta \sin \gamma }\times \frac{{\sin }^2\beta \sin \theta }{\sin \alpha \sin \gamma }\times \frac{{\sin }^2\gamma \sin \theta }{\sin \beta \sin \alpha } \\ \sin (\alpha -\theta )\sin (\beta -\theta )\sin (\gamma -\theta )={\sin }^2\theta \end{gathered}$$