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Prove that co...

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Prove that cos30^°-sin20^°/cos40^°+cos20^°=4/rt3.cos40^°.cos80^°

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Q. Prove that:
(i) cos 10° cos 30° cos 50° cos 70° =

\frac{3}{16}

(ii) cos 40° cos 80° cos 160° =

- \frac{1}{8}

(iii) sin 20° sin 40° sin 80° =

\frac{\sqrt{3}}{8}

(iv) cos 20° cos 40° cos 80° =

\frac{1}{8}

(v) tan 20° tan 40° tan 60° tan 80° = 3

(vi) tan 20° tan 30° tan 40° tan 80° = 1

(vii) sin 10° sin 50° sin 60° sin 70° =

\frac{\sqrt{3}}{16}

(viii) sin 20° sin 40° sin 60° sin 80° =

\frac{3}{16}

Q. Prove that --
cos20°cos40° cos60° cos80° =1

c o s 20^{\circ} c o s 40^{\circ} c o s 60^{\circ} c o s 80^{\circ} = \frac{1}{16}

cos 20^{\circ} \cdot cos 40^{\circ} \cdot cos 60^{\circ} \cdot cos 80^{\circ} = . . . . .

cos 40^{\circ} + cos 50^{\circ} + cos 70^{\circ} + cos 80^{\circ} = cos 20^{\circ} + cos 10^{\circ}

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