## Trigonometry (11th Edition) Clone

By multiplying both numerator and denominator by $(1-\cos A)$, we can derive the equivalent identity as shown below.
$$\tan\frac{A}{2}=\frac{\sin A}{1+\cos A}$$ The job is to derive the equivalent identity: $$\tan\frac{A}{2}=\frac{1-\cos A}{\sin A}$$ As hinted by the exercise, we multiply both numerator and denominator of the original identity by $(1-\cos A)$ $$\tan\frac{A}{2}=\frac{\sin A(1-\cos A)}{(1+\cos A)(1-\cos A)}$$ - Denominator: $$(1+\cos A)(1-\cos A)=1-\cos^2A$$ (as $(A+B)(A-B)=A^2-B^2$) $$(1+\cos A)(1-\cos A)=\sin^2A$$ (Pythagorean Identity) Therefore, $$\tan\frac{A}{2}=\frac{\sin A(1-\cos A)}{\sin^2 A}$$ $$\tan\frac{A}{2}=\frac{1-\cos A}{\sin A}$$ which is the equivalent identity. Thus, we have completed deriving the equivalent identity as requested.