Express the following in terms of angles between 0° and 45° :

sin 59° + tan 63°

Express the following in terms of angles between 0° and 45° :

cosec 68° + cot 72°

For triangle ABC, show that :

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

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

Evaluate :

$\text{3}\dfrac{\text{ sin 72°}}{\text{ cos 18°}}$ - $\dfrac{\text{sec 32°}}{\text{cosec 58°}}$