PROVE THE IDENTITY SIN SQUARE X +COS SQUAREX=1
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As we know sinx= perpendicular /hypotenuse
cosx=base/hypotenuse
then sin²x=AB²/AC²
COS²X=BC²/AC²
then according to question
AB²/AC²+BC²/AC²=1
(AB²+BC²)/AC²=1
AB²+BC²=AC²
according to Pythagoras theurom
hence it proved
were AC = hypotenuse of triangle
AB = PERPENDICULAR of triangle
BC= base of triangle
cosx=base/hypotenuse
then sin²x=AB²/AC²
COS²X=BC²/AC²
then according to question
AB²/AC²+BC²/AC²=1
(AB²+BC²)/AC²=1
AB²+BC²=AC²
according to Pythagoras theurom
hence it proved
were AC = hypotenuse of triangle
AB = PERPENDICULAR of triangle
BC= base of triangle
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