Simplest possible proof of: Sin^2A + Cos^2A = 1.

Consider a right angled triangle ABC, right angled at B.

C
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B__A

SinA = BC/AC
Squaring both sides, we have
Sin^2A = BC^2/AC^2 ----> (1)
CosA = AB/AC
squaring both sides, we have
Cos^2A = AB^2/AC^2 -----> (2)

Adding (1) and (2), we have;

Sin^2A + Cos^2A = BC^2/AC^2 + AB^2/AC^2
Sin^2A + Cos^2A = AC^2/AC^2
(Because: BC^2 AB^2 = AC^2 by Pythagoras Theorem)
Sin^2A + Cos^2A = 1

Hence, proved.

It's simple.
You just have to consider a triangle, say ABC that is right angled at B.

AB is the height, BC is the base and AC is the hypotenuse.

Take the trigonometric ratios ( I am taking angle C), here sinC = AB/AC (height/perpendicular)

cosC = BC/AC ( base/perpendicular)

Square the sinC and cosC terms respective and add them. Also by pythagoras theorem, AB² + BC² = AC²

You'll get the sum to be equla to 1. Try it...

In a right angle triangle, sinA = opposite/hypotenuse
cosA = adjacent/hypotenuse.
Sin^2A + cos^2A = (opposite^2 + adjacent^2)/hypotenuse^2.
By pythagoras = hypotenuse^2/hypotnuse^2 = 1

Proof is trivial. It emerges directly from the Pythagorean Theorem and the definitions of the trig functions.