In a triangle ABC, sec $\dfrac{C + A}{2}$ is equal to :

1. cosec $\dfrac{B}{2}$

2. sec $\dfrac{B}{2}$

3. cosec $\dfrac{B + A}{2}$

4. none of these

$\dfrac{\text{cot 46°}}{\text{tan 44°}} - 3 \dfrac{\text{sec 20°}}{\text{cosec 70°}}$ + 5 is equal to :

1. -3

2. 4

3. 3

4. -4

$\dfrac{\text{cos θ. cos (90° - θ)}}{\text{tan (90° - θ)}}$ is equivalent to :

1. cos² θ - 1

2. sin² θ

3. sin² θ - cos² θ

4. sin² θ - 1

Show that :

tan 10° tan 15° tan 75° tan 80° = 1