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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
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sorry I forgot to attach the image
Okay, I got it, thanks !
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

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