# Pythagorian Identities

Let “OPQ” be a triangle where angle POQ is $\theta$ , and it’s base be “x” and perpendicular “y” as shown in the picture below:

Pythagorian Identities

Then,

Appealing to the Pythagorian theorem, we have:

$OQ^2 + QP^2 = OP^2$

Now let us suppose op be “r” then:

$x^2 + y^2 = r^2$

And as:

$x = r \cos \theta$ and $y = r \sin \theta$

We have:

$r^2 ( \cos ^2 \theta + \sin ^2 \theta = r^2$

Thus: $\sin ^2 \theta + \cos ^2 \theta = 1$

This is called the fundamental Pythagorian identity of trigonometry.

From this we can also develop other identities as:

Dividing both side of fundamental Pythagorian identity by $\cos ^2 \theta$

$\frac{\sin ^2 \theta}{\cos ^2 \theta} + \frac{\cos ^2 \theta}{\cos ^2 \theta} = \frac{1}{\cos ^2 \theta}$

Hence ,$1 + \tan ^2 \theta = \sec ^2 \theta$

And now dividing both side of fundamental Pythagorian identity by $\sin ^2 \theta$

$\frac{\sin ^2 \theta}{\sin ^2 \theta} + \frac{\cos ^2 \theta}{\sin ^2 \theta} = \frac{1}{\sin ^2 \theta}$

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

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