Verify each of the following :

(i) cos 60° cos 30° - sin 60° sin 30° = 0

(ii) cos 60° = (1 - 2 sin²30°) = (2 cos²30° - 1)

(iii) tan 30° = $\Big(\dfrac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ}\Big)$

(i) cos 60° cos 30° - sin 60° sin 30°

= $\dfrac{1}{2}\times \dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{3}}{2}\times\dfrac{1}{2}$

= 0.

Hence, proved that cos 60° cos 30° - sin 60° sin 30° = 0.

(ii) As,

sin² 30° = (sin 30°)² = $\Big(\dfrac{1}{2}\Big)^2= \dfrac{1}{4}$

cos² 30° = (cos 30°)² = $\Big(\dfrac{\sqrt{3}}{2}\Big)^2 = \dfrac{3}{4}$

Therefore

Left Hand Side :

cos 60° = $\dfrac{1}{2}$

Right Hand Side :

(1 - 2 sin²30°)

=1 - 2 $\times \dfrac{1}{4} = \dfrac{1}{2}$

(2 cos²30° - 1)

= 2 $\times\dfrac{3}{4}$ - 1 = $\dfrac{1}{2}$

Hence proved that cos 60° = (1 - 2 sin²30°) = (2 cos²30° - 1).

(iii) Left Hand Side :

tan 30° = $\dfrac{1}{\sqrt{3}}$

Right Hand Side

$\Big(\dfrac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ}\Big) = \dfrac{\sqrt{3} - \dfrac{1}{\sqrt{3}}}{1 + \sqrt{3}\times\dfrac{1}{\sqrt{3}}}\\[1em] =\dfrac{\dfrac{3-1}{\sqrt{3}}}{1 + 1}\\[1em] =\dfrac{\dfrac{2}{\sqrt{3}}}{2}\\[1em] = \dfrac{1}{\sqrt{3}}.$

Hence proved that tan 30° = $\Big(\dfrac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ}\Big)$