### Theory:

Consider an isosceles triangle $$ABC$$ with angle measure $$45^{\circ}$$ right-angled at $$B$$.

Let the measure of the equal sides of the triangle be $$a$$ units.

First, let us calculate the measure of the hypotenuse in the figure.

Since the given triangle is a right-angled triangle by the Pythagoras theorem, we have:
In a right angled triangle, $$\text{Hypotenuse}^{2} = \text{Adjacent side}^{2} + \text{Opposite side}^{2}$$.
$$AC^2$$ $$=$$ $$AB^2$$ $$+$$ $$BC^2$$.

$$AC^2$$ $$=$$ $$a^2$$ $$+$$ $$a^2$$

$$AC^2$$ $$=$$  $$2a^2$$

$$\Rightarrow AC$$ $$=$$ $$\sqrt{2a^2}$$

$$AC$$ $$=$$ $$a \sqrt{2}$$ units

Therefore, for the given right-angled isosceles triangle we have:

Opposite side $$=$$ $$a$$ units

Adjacent side $$=$$ $$a$$ units

Hypotenuse $$=$$ $$a \sqrt{2}$$ units

Now, let us determine all the trigonometric ratios of $$45^{\circ}$$.

• Sine $$45^{\circ}$$:

$$\sin 45^{\circ}$$ $$=$$ $$\frac{\text{Opposite side}}{\text{Hypotenuse}}$$

$$=$$ $$\frac{a}{a \sqrt{2}}$$

$$=$$ $$\frac{1}{\sqrt{2}}$$

• Cosine $$45^{\circ}$$:

$$\cos 45^{\circ}$$ $$=$$ $$\frac{\text{Adjacent side}}{\text{Hypotenuse}}$$

$$=$$ $$\frac{a}{a \sqrt{2}}$$

$$=$$ $$\frac{1}{\sqrt{2}}$$

• Tangent $$45^{\circ}$$:

$$\tan 45^{\circ}$$ $$=$$ $$\frac{\text{Opposite side}}{\text{Adjacent side}}$$

$$=$$ $$\frac{a}{a}$$

$$=$$ $$1$$

Using these basic trigonometric ratios determine their reciprocals as follows:

• Cosecant $$45^{\circ}$$:

$$\text{cosec}\,\theta$$ $$=$$ $$\frac{1}{\sin \theta}$$

$$=$$ $$\sqrt{2}$$

• Secant $$45^{\circ}$$:

$$\sec \theta$$ $$=$$ $$\frac{1}{\cos \theta}$$

$$=$$ $$\sqrt{2}$$

• Cotangent $$45^{\circ}$$:

$$\cot \theta$$ $$=$$ $$\frac{1}{\tan \theta}$$

$$=$$ $$1$$

Let us summarize all the trigonometric ratios of $$45^{\circ}$$ in the following table.

 $$\sin \theta$$ $$\cos \theta$$ $$\tan \theta$$ $$\text{cosec}\,\theta$$ $$\sec \theta$$ $$\cot \theta$$ $$\theta = 45^{\circ}$$ $$\frac{1}{\sqrt{2}}$$ $$\frac{1}{\sqrt{2}}$$ $$1$$ $$\sqrt{2}$$ $$\sqrt{2}$$ $$1$$