SOLUTION: Make the trigonometric substitution x = a sec θ for 0 < θ < π/2 and a > 0. Simplify the resulting expression. {{{sqrt(x^2-a^2)}}}/x I got asinθ which is wrong. Where did

Algebra ->  Trigonometry-basics -> SOLUTION: Make the trigonometric substitution x = a sec θ for 0 < θ < π/2 and a > 0. Simplify the resulting expression. {{{sqrt(x^2-a^2)}}}/x I got asinθ which is wrong. Where did       Log On


   

Question 1159664: Make the trigonometric substitution
x = a sec θ for 0 < θ < π/2 and a > 0.
Simplify the resulting expression.
sqrt%28x%5E2-a%5E2%29/x
I got asinθ which is wrong. Where did I go wrong?
This is my work:
sqrt%28x%5E2-a%5E2%29/x =
sqrt%28%28a%5E2%28sec%5E2%CE%B8-1%29%29/2 =
sqrt%28%28a%5E2%28tan%5E2%CE%B8%29/asecθ =
=atan/asecθ
=asin/cosθx(cosθ/1)
= asinθ
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%28x%5E2-a%5E2%29%2Fx

sqrt%28%28a%2Asec%28theta%29%29%5E2-a%5E2%29%2F%28a%2Asec%28theta%29%29

sqrt%28a%5E2%2Asec%5E2%28theta%29-a%5E2%29%2F%28a%2Asec%28theta%29%29

sqrt%28a%5E2%28sec%5E2%28theta%29-1%29%29%2F%28a%2Asec%28theta%29%29

a%2Asqrt%28tan%5E2%28theta%29%29%2F%28a%2Asec%28theta%29%29

%28cross%28a%29%2Atan%28theta%29%29%2F%28cross%28a%29%2Asec%28theta%29%29


tan%28theta%29%2Fsec%28theta%29.........tan%28theta%29=sin%28theta%29%2Fcos%28theta%29, and sec%28theta%29=1%2Fcos%28theta%29


%28sin%28theta%29%2Fcos%28theta%29%29%2F%281%2Fcos%28theta%29%29


%28sin%28theta%29%2Fcross%28cos%28theta%29%291%29%2F%281%2Fcross%28cos%28theta%29%291%29


sin%28theta%29