Show that :

sin 42° sec 48° + cos 42° cosec 48° = 2

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° :

cos 74° + sec 67°

For triangle ABC, show that :

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

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