## Trigonometry (11th Edition) Clone

$$\tan\frac{\theta}{2}=\sqrt{\frac{b}{2R-b}}$$
- From half-angle identity for tangent: $$\tan\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$$ The angle $\frac{\theta}{2}$ corresponds to the angle in the image. The angle in the image is definitely greater than $0^\circ$ and smaller than $90^\circ$, so it lies in quadrant I, where tangents are positive. Therefore, $\tan\frac{\theta}{2}\gt0$. We need to select positive square root. $$\tan\frac{\theta}{2}=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$$ - From part a), we found that $$\cos\frac{\theta}{2}=\frac{R-b}{R}$$ Replace the formula of $\cos\frac{\theta}{2}$ into the formula of $\tan\frac{\theta}{2}$: $$\tan\frac{\theta}{2}=\sqrt{\frac{1-\frac{R-b}{R}}{1+\frac{R-b}{R}}}$$ $$\tan\frac{\theta}{2}=\sqrt{\frac{\frac{R-R+b}{R}}{\frac{R+R-b}{R}}}$$ $$\tan\frac{\theta}{2}=\sqrt{\frac{\frac{b}{R}}{\frac{2R-b}{R}}}$$ $$\tan\frac{\theta}{2}=\sqrt{\frac{b}{2R-b}}$$