Question

How to prove cos^2theta + sin^2theta = 1 ?
Please no short answers, a good answer with explanation ! 

Answer 1

This is very simple, in the image attached, you can see for the angle theta in the right angled triangle, AC is the p [perpendicular], BC is the b [base] and AB is h [hypotenuse].
Cos^2theta + sin^2 theta can be written as :-
$\frac{ b^{2} }{ h^{2} } + \frac{ p^{2} }{ h^{2} }$..........[because cos theta = b/h and sin theta = p/h]
So, since b^2 + p^2 = h^2 [pythagoras theorem], we can write $\frac{ b^{2} }{ h^{2} } + \frac{ p^{2} }{ h^{2} }$ as h^2/h^2 = 1. 
Hence proved

Answer 2

sin α =$\frac{perpendicular}{hypotnuse} = \frac{p}{h}$

cos α=$\frac{base}{hypotnuse}= \frac{b}{h}$

$sin^{2} \alpha =\frac{{P^{2}}}{h^{2}}$

$cos^{2} \alpha = \frac{b^{2} }{h^{2} }$

$sin^{2} \alpha + cos^{2} \alpha = \frac{p^{2} }{h^{2} }+ \frac{b^{2} }{h^{2} }= \frac{p^{2} +b^{2} }{h^{2} }= \frac{h^{2} }{h^{2} } =1$

becuse acc. to pythagorus theorm $p^{2} +b^{2} =h^{2}$

hence $sin^{2} \alpha +cos^{2} \alpha =1$