If A, B and C are the interior angles of a △ABC, show that :

(i) cos $\dfrac{A + B}{2}$ = sin $\dfrac{C}{2}$

(ii) tan $\dfrac{C + A}{2}$ = cot $\dfrac{B}{2}$

(i) Given,

A, B and C are the interior angles of a △ABC.

∴ A + B + C = 180°

⇒ $\dfrac{A + B + C}{2} = 90°$

⇒ $\dfrac{A + B}{2} = 90° - \dfrac{C}{2}$

To prove,

cos $\dfrac{A + B}{2}$ = sin $\dfrac{C}{2}$

Substituting value of $\dfrac{A + B}{2}$ in L.H.S. of the equation we get :

$\phantom{=} \text{cos } \Big(\dfrac{A + B}{2}\Big) \\[1em] = \text{cos } \Big(90\degree - \dfrac{C}{2}\Big) \\[1em] = \text{sin } \dfrac{C}{2} \space [\because \text{cos (90 - θ)} = \text{sin θ}]$

Since, L.H.S. = R.H.S.

Hence, proved that cos $\dfrac{A + B}{2}$ = sin $\dfrac{C}{2}$.

(ii) Given,

A, B and C are the interior angles of a △ABC.

∴ A + B + C = 180°

⇒ $\dfrac{A + B + C}{2} = 90°$

⇒ $\dfrac{A + C}{2} = 90° - \dfrac{B}{2}$

To prove,

tan $\dfrac{C + A}{2}$ = cot $\dfrac{B}{2}$

Substituting value of $\dfrac{C + A}{2}$ in L.H.S. of equation we get :

$\phantom{=} \text{tan } \Big(\dfrac{C + A}{2}\Big) \\[1em] = \text{tan } \Big(90\degree - \dfrac{B}{2}\Big) \\[1em] = \text{cot } \dfrac{B}{2} \space [\because \text{tan (90 - θ)} = \text{cot θ}]$

Since, L.H.S. = R.H.S.

Hence, proved that tan $\dfrac{C + A}{2}$ = cot $\dfrac{B}{2}$.