The subtraction of square of cotangent function from cosecant function at an angle equals to one, is called Pythagorean identity for cosecant and cotangent functions.

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

$\overline{BC}$, $\overline{AB}$ and $\overline{AC}$ are the opposite side, adjacent side and hypotenuse of the $\Delta CAB$ respectively and the associated lengths are $BC$, $AB$ and $AC$.

Write the mathematical relation between three sides of the right angled triangle as per Pythagorean Theorem.

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

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

Do an adjustment acceptably to express the equation in terms of cosecant and cotangent functions. It can be done by dividing both sides of the equation by the square of length of the side $\overline{BC}$.

$\implies \dfrac{{BC}^2 + {AB}^2}{{BC}^2} = \dfrac{{AC}^2}{{BC}^2}$

$\implies \dfrac{{BC}^2}{{BC}^2} + \dfrac{{AB}^2}{{BC}^2} = \dfrac{{AC}^2}{{BC}^2}$

$\implies \Bigg(\dfrac{BC}{BC}\Bigg)^2 + \Bigg(\dfrac{AB}{BC}\Bigg)^2 = \Bigg(\dfrac{AC}{BC}\Bigg)^2$

$\implies 1 + \Bigg(\dfrac{AB}{BC}\Bigg)^2 = \Bigg(\dfrac{AC}{BC}\Bigg)^2$

Now, express the ratios of sides in terms of trigonometric functions according to the $\Delta BAC$.

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

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

Substitute the ratios of the sides by the associated trigonometric functions.

$\implies 1 + (\cot{\theta})^2 = (\csc{\theta})^2$

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

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

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

Therefore, it is proved that the subtraction of square of cotangent function from square of cosecant function at an angle is equal to one. This trigonometric identity is derived according to the Pythagoras Theorem. Hence, it is generally known as the Pythagorean identity for cosecant and cotangent functions.

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