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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. 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 $$=$$ $$1$$ units Opposite side $$=$$ $$\sqrt{3}$$ units Adjacent side $$=$$ $$\sqrt{3}$$ units Adjacent side $$=$$ $$1$$ units Hypotenuse $$=$$ $$2$$ units Hypotenuse $$=$$ $$2$$ 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.

 $$\sin \theta$$ $$\cos \theta$$ $$\tan \theta$$ $$\text{cosec}\,\theta$$ $$\sec \theta$$ $$\cot \theta$$ $$\theta = 30^{\circ}$$ $$\frac{1}{2}$$ $$\frac{\sqrt{3}}{2}$$ $$\frac{1}{\sqrt{3}}$$ $$2$$ $$\frac{2}{\sqrt{3}}$$ $$\sqrt{3}$$ $$\theta = 60^{\circ}$$ $$\frac{\sqrt{3}}{2}$$ $$\frac{1}{2}$$ $$\sqrt{3}$$ $$\frac{2}{\sqrt{3}}$$ $$2$$ $$\frac{1}{\sqrt{3}}$$