Subtraction of square of cotangent from square of cosecant at an angle equals to $1$ is called Pythagorean trigonometric identity of cosecant and cotangent.

The relation of cosecant with cotangent is developed mathematically by using Pythagorean Theorem. Hence, it is called as Pythagorean trigonometric identity and also called as Pythagoras trigonometric identity in trigonometry.

It is used mainly to express cosecant in terms of cotangent and also used to express cotangent in terms of cosecant.

$\Delta BAC$ is a right angled triangle and angle of right angled triangle is assumed theta $(\theta)$.

Use Pythagorean Theorem and apply it to this triangle.

${AC}^2 = {BC}^2 + {AB}^2$

$\Rightarrow {BC}^2 + {AB}^2 = {AC}^2$

Divide left hand side and right hand side expressions by ${BC}^2$.

$$\Rightarrow \frac{{BC}^2 + {AB}^2}{{BC}^2} = \frac{{AC}^2}{{BC}^2}$$

$$\Rightarrow \frac{{BC}^2}{{BC}^2} + \frac{{AB}^2}{{BC}^2} = \frac{{AC}^2}{{BC}^2}$$

$$\Rightarrow \Bigg(\frac{BC}{BC}\Bigg)^2 + \Bigg(\frac{AB}{BC}\Bigg)^2 = \Bigg(\frac{AC}{BC}\Bigg)^2$$

$$\Rightarrow 1 + \Bigg(\frac{AB}{BC}\Bigg)^2 = \Bigg(\frac{AC}{BC}\Bigg)^2$$

According to right angled triangle $BAC$

$$\frac{AB}{BC} = \cot \theta$$

$$\frac{AC}{BC} = \csc \theta$$

Substitute these two ratios in terms of associated trigonometric functions.

$\Rightarrow 1 + (\cot \theta)^2 = (\csc \theta)^2$

$\Rightarrow 1 + \cot^2 \theta = \csc^2 \theta$

$\Rightarrow 1 = \csc^2 \theta \, – \, \cot^2 \theta$

$\Rightarrow \csc^2 \theta \, – \, \cot^2 \theta = 1$

It is proved that cosecant squared angle minus cotangent squared angle is equal to $1$. It is proved from Pythagoras theorem. So, it is known as a Pythagorean trigonometric identity in trigonometry.

Assume angle of triangle $\theta = 60^°$ to verify this trigonometric identity.

$$\csc^2 \theta \, – \, \cot^2 \theta = \csc^2 60^° \, – \, \cot^2 60^°$$

$$\Rightarrow \csc^2 60^° \, – \, \cot^2 60^° = \Bigg(\frac{2}{\sqrt{3}}\Bigg)^2 \,- \Bigg(\frac{1}{\sqrt{3}}\Bigg)^2$$

$$\Rightarrow \csc^2 60^° \, – \, \cot^2 60^° = \frac{4}{3} \,-\, \frac{1}{3}$$

$$\Rightarrow \csc^2 60^° \, – \, \cot^2 60^° = \frac{4 \,-\, 1}{3}$$

$$\Rightarrow \csc^2 60^° \, – \, \cot^2 60^° = \frac{3}{3}$$

$$\Rightarrow \csc^2 60^° \, – \, \cot^2 60^° = 1$$

It is mathematically proved that subtraction of square of cotangent from square of cosecant at angle $60^°$ is equal to $1$. It is also proved for every angle. Hence, the mathematical relation between cosecant and cotangent is called as trigonometric identity.

The Pythagoras trigonometric identity of cotangent and cosecant can also be used to express cosecant in terms of cotangent and also used to express cotangent in terms of cosecant.

1

Cosecant can be written in terms of cotangent in two different forms.

$$\csc^2 \theta = 1 + \cot^2 \theta$$

$$\csc \theta = \pm \sqrt{1 + \cot^2 \theta}$$

2

Cotangent can be written in terms of cosecant in two different forms.

$$\cot^2 \theta = \csc^2 \theta -1$$

$$\cot \theta = \pm \sqrt{\csc^2 \theta \,-\, 1}$$