Theory:

Consider an equilateral triangle \(ABC\) with sides measuring \(2\) units.
 
That is \(AB\) \(=\) \(BC\) \(=\) \(CA\) \(=\) \(2\) units.
 
Draw a bisector of \(\angle A\) such that it meets \(BC\) at \(D\).
 
The angle bisector of an equilateral triangle also bisects the side opposite it.
 
So, \(BD\) \(=\) \(DC\) \(=\) \(1\) unit.
 
30 60 deg.png
 
First, let us calculate the measure of angle bisector \(BD\) in the figure.
 
Consider the triangle \(ABD\).
 
Since the given triangle is a right-angled triangle by the Pythagoras theorem, we have:
In a right angled triangle, \(\text{Hypotenuse}^{2} = \text{Adjacent side}^{2} + \text{Opposite side}^{2}\).
\(AB^2\) \(=\) \(BD^2\) \(+\) \(DA^2\).
 
\(DA^2\) \(=\) \(AB^2\) \(-\) \(BD^2\)
 
\(DA^2\) \(=\)  \(2^2\) \(-\) \(1^2\)
 
\(DA^2\) \(=\) \(4 - 1\)
 
\(DA^2\) \(=\) \(3\)
 
\(\Rightarrow DA\) \(=\) \(\sqrt{3}\)
 
Therefore,  for the considered right-angled triangle, we have:
 
With respect to \(30^{\circ}\)With respect to \(60^{\circ}\)
Opposite side \(=\) \(1\) unitsOpposite side \(=\) \(\sqrt{3}\) units
Adjacent side \(=\) \(\sqrt{3}\) unitsAdjacent side \(=\) \(1\) units
Hypotenuse \(=\) \(2\) unitsHypotenuse \(=\) \(2\) units
 
Now, let us determine all the trigonometric ratios of \(30^{\circ}\) and \(60^{\circ}\).
 
  • Sine:
 
\(\sin 30^{\circ}\)
\(\sin 60^{\circ}\)
\(\sin 30^{\circ}\) \(=\) \(\frac{\text{Opposite side}}{\text{Hypotenuse}}\)
 
\(=\) \(\frac{1}{2}\)
\(\sin 60^{\circ}\) \(=\) \(\frac{\text{Opposite side}}{\text{Hypotenuse}}\)
 
\(=\) \(\frac{\sqrt{3}}{2}\)
 
  • Cosine \(30^{\circ}\):
 
\(\cos 30^{\circ}\)
\(\cos 60^{\circ}\)
\(\cos 30^{\circ}\) \(=\) \(\frac{\text{Adjacent side}}{\text{Hypotenuse}}\)
 
\(=\) \(\frac{\sqrt{3}}{2}\)
\(\cos 60^{\circ}\) \(=\) \(\frac{\text{Adjacent side}}{\text{Hypotenuse}}\)
 
\(=\) \(\frac{1}{2}\)
 
  • Tangent:
 
\(\tan 30^{\circ}\)
\(\tan 60^{\circ}\)
\(\tan 30^{\circ}\) \(=\) \(\frac{\text{Opposite side}}{\text{Adjacent side}}\)
 
\(=\) \(\frac{1}{\sqrt{3}}\)
\(\tan 60^{\circ}\) \(=\) \(\frac{\text{Opposite side}}{\text{Adjacent side}}\)
 
\(=\) \(\frac{\sqrt{3}}{1}\)
 
\(=\) \(\sqrt{3}\)
 
Using these basic trigonometric ratios determine their reciprocals as follows:
 
  • Cosecant:
 
\(\text{cosec}\,30^{\circ}\)
\(\text{cosec}\,60^{\circ}\)
\(\text{cosec}\,30^{\circ}\) \(=\) \(\frac{1}{\sin 30^{\circ}}\)
 
\(=\) \(\frac{2}{1}\)
 
\(=\) \(2\)
\(\text{cosec}\,60^{\circ}\) \(=\) \(\frac{1}{\sin 60^{\circ}}\)
 
\(=\) \(\frac{2}{\sqrt{3}}\)
 
  • Secant:
 
\(\sec 30^{\circ}\)
\(\sec 60^{\circ}\)
\(\sec 30^{\circ}\) \(=\) \(\frac{1}{\cos 30^{\circ}}\)
 
\(=\) \(\frac{2}{\sqrt{3}}\)
\(\sec 60^{\circ}\) \(=\) \(\frac{1}{\cos 60^{\circ}}\)
 
\(=\) \(\frac{2}{1}\)
 
\(=\) \(2\)
 
  • Cotangent:
 
\(\cot 30^{\circ}\)
\(\cot 60^{\circ}\)
\(\cot 30^{\circ}\) \(=\) \(\frac{\text{1}}{\tan 30^{\circ}}\)
 
\(=\)  \(\frac{\sqrt{3}}{1}\)
 
\(=\) \(\sqrt{3}\)
\(\cot 60^{\circ}\) \(=\) \(\frac{\text{1}}{\tan 60^{\circ}}\)
 
\(=\) \(\frac{1}{\sqrt{3}}\)
 
Let us summarize all the trigonometric ratios of \(30^{\circ}\) and \(60^{\circ}\) in the following table.
 
 
\(\sin \theta\)
\(\cos \theta\)
\(\tan \theta\)
\(\text{cosec}\,\theta\)
\(\sec \theta\)
\(\cot \theta\)
\(\theta = 30^{\circ}\)
\(\frac{1}{2}\)
\(\frac{\sqrt{3}}{2}\)
\(\frac{1}{\sqrt{3}}\)
\(2\)
\(\frac{2}{\sqrt{3}}\)
\(\sqrt{3}\)
\(\theta = 60^{\circ}\)
\(\frac{\sqrt{3}}{2}\)
\(\frac{1}{2}\)
\(\sqrt{3}\)
\(\frac{2}{\sqrt{3}}\)
\(2\)
\(\frac{1}{\sqrt{3}}\)