Question

21

Answer 1

Answer:

Step-by-step explanation:

Make the substitution first for x:

$\frac{5(2sint)}{\sqrt{4-(2sint)^2} }$ and simplify a bit:

$\frac{10sint}{\sqrt{4-4sin^2t} }$. Factor a 4 out of the expression under the radical:

$\frac{10sint}{\sqrt{4(1-sin^2t)} }$ and pull out the perfect square in 4 as a 2:

$\frac{10sint}{2\sqrt{1-sin^2t} }$ and divide 10 by 2 to get:

$\frac{5sint}{\sqrt{1-sin^2t} }$

Our Pythagorean trig identity tells us that

$sin^2(t)+cos^2(t)=1$ so

$1-sin^2(t)=cos^2(t)$. Make that substitution:

$\frac{5sint}{\sqrt{cos^2t} }$  The denominator can be simplified (the squaring of the square root each cancel out), leaving us with:

$\frac{5sint}{cost}$.  Sin over cos is the same as tangent, so our final simplification is

5tan(t)