Question

2x=3-4 tan theta, 3y=5+3 sec theta eliminate theta

Answer 1

Step-by-step explanation:

see square both the equations and add it you will get your and.

equation 1: 2x=3-4tan(thieta)

4tan(thieta)=3-2x

tan(thieta)=(3-2x)/4

Now on squaring it we get ,

tan^2(thieta)=(3-2x)^2/4^2 ------(1)

Equation 2:

3y=5+3sec(thieta)

3sec(thieta)=3y-5

sec(thieta)= (3y-5)/3

Now in squaring it we get,

sec^2(thieta)=(3y-5)^2/(3^2)-------(2)

Now substracting equation 1 from equation 2 we get ,

sec^2(thieta) - tan^2(thieta)=

((3y-5)^2/9) - ((3-2x)^2/16)

therefore you get,

1=((3y-5)^2/9 ) - ((3-2x)^2/16)

since, sec^2(thieta)-tan^2(thieta)= 1

now , thieta is eliminated and you got a simultaneous quadratic equation. Which you can also use to find the value of X and Y

Answer 2

Step-by-step explanation: