For triangle ABC, show that :

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

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

(i) In triangle ABC,

⇒ ∠A + ∠B + ∠C = 180° [By angle sum property of triangle]

⇒ ∠A + ∠B = 180° - ∠C ………(1)

Given equation,

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

Substituting value of (A + B) from (1) in L.H.S. of above equation :

$\Rightarrow \text{sin} \dfrac{180° - C}{2} \\[1em] \Rightarrow \text{sin} \Big(90° - \dfrac{C}{2}\Big)$

By formula,

sin(90° - θ) = cos θ

$\therefore \text{cos} \dfrac{C}{2}$.

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

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

(ii) In triangle ABC,

⇒ ∠A + ∠B + ∠C = 180° [By angle sum property of triangle]

⇒ ∠B + ∠C = 180° - ∠A ………(1)

Given equation,

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

Substituting value of (B + C) from (1) in L.H.S. of above equation :

$\Rightarrow \text{tan} \dfrac{180° - A}{2} \\[1em] \Rightarrow \text{tan } \Big(90° - \dfrac{A}{2}\Big)$

By formula,

tan(90° - θ) = cot θ

$\Rightarrow \text{cot} \dfrac{A}{2}$.

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

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