Question 1159331
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We are given {{{x = a*tan(theta)}}}


Wherever you see an 'x', replace it with {{{a*tan(theta)}}}. Then simplify.


{{{y = (x^2)/(sqrt(a^2 + x^2))}}}


{{{y = ((a*tan(theta))^2)/(sqrt(a^2 + (a*tan(theta))^2))}}} Make the replacements


{{{y = (a^2*tan^2(theta))/(sqrt(a^2 + a^2*tan^2(theta)))}}} 


{{{y = (a^2*tan^2(theta))/(sqrt(a^2*1 + a^2*tan^2(theta)))}}}


{{{y = (a^2*tan^2(theta))/(sqrt(a^2(1 + tan^2(theta))))}}} Factor out the a^2


{{{y = (a^2*tan^2(theta))/(sqrt(a^2)*sqrt((1 + tan^2(theta))))}}} Break up the root


{{{y = (a^2*tan^2(theta))/(a*sqrt(1 + tan^2(theta)))}}} Works because a > 0


{{{y = (a*tan^2(theta))/(sqrt(1 + tan^2(theta)))}}}


{{{y = (a*tan^2(theta))/(sqrt(sec^2(theta)))}}} The equation 1+tan^2 = sec^2 is one variation of the pythagorean trig identity


{{{y = (a*tan^2(theta))/(sec(theta)))}}} Secant is positive when theta is between -pi/2 and pi/2


{{{y = a*tan^2(theta)*((1)/(sec(theta)))}}}


{{{y = a*tan^2(theta)*cos(theta)}}}


{{{y = a*((sin^2(theta))/(cos^2(theta)))*cos(theta)}}}


{{{y = a*((sin^2(theta))/(cos(theta)*cos(theta)))*cos(theta)}}}


{{{y = a*((sin^2(theta))/(cos(theta)*cross(cos(theta))))*cross(cos(theta))}}} One pair of cosine terms divide and cancel


{{{y = (a*sin^2(theta))/(cos(theta))}}}


{{{y = (a*sin(theta)*sin(theta))/(cos(theta))}}}


{{{y = a*sin(theta)*((sin(theta))/(cos(theta)))}}}


{{{y = a*sin(theta)*tan(theta))}}} 


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Since {{{y = (x^2)/(sqrt(a^2 + x^2))}}} and {{{y = a*sin(theta)*tan(theta))}}}, we can say


{{{(x^2)/(sqrt(a^2 + x^2)) = a*sin(theta)*tan(theta))}}} only when a > 0 and -pi/2 < theta < pi/2.
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