show that equality $\displaystyle\frac{1}{a}-\frac{1}{b}-\frac{1}{c}=0$ hold when a,b,c are triangle sides

Question

show that equality $\displaystyle\frac{1}{a}-\frac{1}{b}-\frac{1}{c}=0$ hold when a,b,c are triangle sides which has angles: $\alpha, 2\alpha, 4\alpha$

I tired to do it using sines law but didn't work

Answer 1

If $D+2D+4D=\pi$

$$\frac1{\sin2D}+\frac1{\sin4D}=\frac{\sin2D+\sin4D}{2\sin2D\sin4D}$$

Using Prosthaphaeresis & Double-Angle Formulas, this equals

$$\frac{2\sin3D\cos D}{2\sin D\cos D\sin4D}=\frac1{\sin D}$$

as $\sin3D=\sin(\pi-4D)=\sin4D$ and the cancellation of $\cos D$ is legitimate as $\displaystyle D=\frac\pi7\implies\cos D=\cos\frac\pi7\ne0$

Now use Law of Sines