Question 923776
(f º g)(t) is the same as {{{f(g(t))}}}



{{{0 < t < pi/2}}} which means t is in quadrant I



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{{{f(x) = (4x)/(sqrt(1-x^2))}}} Start with f(x)



{{{f(g(t)) = (4*g(t))/(sqrt(1-(g(t))^2))}}} Replace every x with g(t)



{{{f(g(t)) = (4*sin(t))/(sqrt(1-(sin(t))^2))}}} Replace every g(t) on the right side with {{{sin(t)}}}



{{{f(g(t)) = (4*sin(t))/(sqrt((cos(t))^2))}}} Use a trig identity



{{{f(g(t)) = (4*sin(t))/(cos(t))}}} Take the square root. t is in quadrant I, so {{{cos(t)}}} is positive.


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Let me know if that helps or not. Thanks.


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