# Solving Pythagorean Identities

The Pythagorean identities in trigonometry are the three identities that come from the Pythagorean theorem.

Recall that the Pythagorean theorem states that the hypotenuse squared of a right triangle is the sum of the square of each of the other two sides.
a

Where c stands for the hypotenuse, and a and b are other two sides of the right triangle. From this theorem, three identities can be determined from substituting in sine and cosine as follows:

sin

tan

1 + cot

^{2}+ b^{2}= c^{2}Where c stands for the hypotenuse, and a and b are other two sides of the right triangle. From this theorem, three identities can be determined from substituting in sine and cosine as follows:

sin

^{2}θ + cos^{2}θ = 1tan

^{2}θ + 1 = sec^{2}θ1 + cot

^{2}θ = cosec^{2}θ**Proof of Pythagorean Identities :**

Lets drow an unit circle as showing in picture and draw an angle θ since it is a unit circle so line CP = 1, let draw the perpendicual lines to x and y axis as PN and PM.

As we know sinθ = opposite-side(O) / hypotunse and cosθ = adjacent-side (A) / hypotunuse

since hypotunuse = radius = 1 herefore we can write as follows:

sinθ = O and cosθ = A

We know the Pythagorean theorem as CP

^{2}= PN

^{2}+ CN

^{2}or 1

^{2}= O

^{2}+ A

^{2}

Now replace the value of O and A with sinθ and cosθ; so we will get

1

^{2}= (sinθ)

^{2}+ (cosθ)

^{2}

1 = sin

^{2}θ + cos

^{2}θ

sin

^{2}θ + cos

^{2}θ = 1 --------(1)

To prove other two indentities we can use following formulas:

1 / sinθ = cscθ

1 / cosθ = secθ

sinθ / cosθ = tanθ

cosθ / sinθ = cotθ

let take the basic indentity 1 = sin

^{2}θ + cos

^{2}θ and device both side by cos

^{2}θ

1 / cos

^{2}θ = sin

^{2}θ / cos

^{2}θ + cos

^{2}θ / cos

^{2}θ

1 / cos

^{2}θ = sin

^{2}θ / cos

^{2}θ + 1

apply basic formulas for 1 / cosθ = secθ and sinθ / cosθ = tanθ, then we will get

sec

^{2}θ = tan

^{2}θ + 1

1 + tan

^{2}θ = sec

^{2}θ -------------(2)

again take the basic indentity 1 = sin

^{2}θ + cos

^{2}θ and device both side by sin

^{2}θ

1 / sin

^{2}θ = sin

^{2}θ / sin

^{2}θ + cos

^{2}θ / sin

^{2}θ

1 / sin

^{2}θ = 1 + cos

^{2}θ / sin

^{2}θ

apply basic formulas for 1 / sinθ = cscθ and cosθ / sinθ = cotθ, then we will get

csc

^{2}θ = 1 + cot

^{2}θ

1 + cot

^{2}θ = csc

^{2}θ -------------(3)