$\tan\left(\frac{A}{2}\right) = \frac{\sin(A/2)}{\cos(A/2)} = \frac{\sqrt{\frac{(s-b)(s-c)}{bc}}}{\sqrt{\frac{s(s-a)}{bc}}} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$

Similarly, formulae for $\sin(B/2)$, $\cos(B/2)$, $\tan(B/2)$, $\sin(C/2)$, $\cos(C/2)$, $\tan(C/2)$ can be derived by cyclic permutation.

$\sin\left(\frac{B}{2}\right) = \sqrt{\frac{(s-a)(s-c)}{ac}}$
$\cos\left(\frac{B}{2}\right) = \sqrt{\frac{s(s-b)}{ac}}$
$\tan\left(\frac{B}{2}\right) = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$
$\sin\left(\frac{C}{2}\right) = \sqrt{\frac{(s-a)(s-b)}{ab}}$
$\cos\left(\frac{C}{2}\right) = \sqrt{\frac{s(s-c)}{ab}}$
$\tan\left(\frac{C}{2}\right) = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$

3. Coplanarity of Vectors

Theorem:

If $\vec{a}$ and $\vec{b}$ are two non-zero and non-collinear vectors, then any vector $\vec{r}$ coplanar with $\vec{a}$ and $\vec{b}$ can be expressed as $\vec{r} = t_1\vec{a} + t_2\vec{b}$, where $t_1$ and $t_2$ are unique scalars.

Proof:

Only If-part:

Assume $\vec{r}$ is coplanar with $\vec{a}$ and $\vec{b}$. Let $\vec{a}$ be along $OA$ and $\vec{b}$ along $OB$. Given a vector $\vec{r}$ with initial point $O$, let $OP=\vec{r}$. Draw lines parallel to $OB$ meeting $OA$ in $M$ and parallel to $OA$ meeting $OB$ in $N$.

Then $\vec{ON} = t_2\vec{b}$ and $\vec{OM} = t_1\vec{a}$ for some $t_1, t_2 \in \mathbb{R}$. By triangle law or parallelogram law, we have $\vec{r} = t_1\vec{a} + t_2\vec{b}$.

If-part:

Suppose $\vec{r} = t_1\vec{a} + t_2\vec{b}$. Since $\vec{a}$ and $\vec{b}$ are coplanar, $t_1\vec{a}$ and $t_2\vec{b}$ are also coplanar. Therefore, their sum $t_1\vec{a} + t_2\vec{b} = \vec{r}$ is also coplanar with $\vec{a}$ and $\vec{b}$.

Uniqueness:

Suppose $\vec{r}$ can also be written as $\vec{r} = s_1\vec{a} + s_2\vec{b}$.

Then $t_1\vec{a} + t_2\vec{b} = s_1\vec{a} + s_2\vec{b}$.

$(t_1-s_1)\vec{a} + (t_2-s_2)\vec{b} = \vec{0}$.

Since $\vec{a}$ and $\vec{b}$ are non-collinear and non-zero, this implies that the coefficients must be zero:

$t_1-s_1=0 \Rightarrow t_1=s_1$

$t_2-s_2=0 \Rightarrow t_2=s_2$

Thus, the scalars $t_1, t_2$ are unique.

4. Condition for Coplanarity of Three Vectors

Theorem:

Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if and only if there exists a non-zero linear combination $x\vec{a} + y\vec{b} + z\vec{c} = \vec{0}$ with $(x,y,z) \neq (0,0,0)$.

Proof:

Only If-part:

Assume $\vec{a}, \vec{b}, \vec{c}$ are coplanar.

Case 1: Suppose any two of $\vec{a}, \vec{b}, \vec{c}$ are collinear, say $\vec{a}$ and $\vec{b}$.

Then there exist scalars $x, y$