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sin@cos@tan@=1-cos^2@ to be provedp=pependicularb=baseh=hypotenuse, therefore the above will bep/h×b/h×p/b=1-b^2/h^2p^2/h^2=1-b^2/h^2now, 1-b^2/h^2=(h^2-b^2)/h^2by phytogores therom h^2-b^2=p^2hence left hand side=right hand side
Amazing stuff ❤
Thank you
New friend
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Wote am becoming. Better
sin@cos@tan@=1-cos^2@ to be proved
p=pependicular
b=base
h=hypotenuse, therefore the above will be
p/h×b/h×p/b=1-b^2/h^2
p^2/h^2=1-b^2/h^2
now, 1-b^2/h^2=(h^2-b^2)/h^2
by phytogores therom h^2-b^2=p^2
hence left hand side=right hand side
Amazing stuff ❤
Thank you
New friend
❤
Wote am becoming. Better