# Problems on Eliminate Theta

Here we will solve various types of problems on eliminate theta from the given equations.

We know, “eliminate theta from the equations” means that the equations are combined in such a way into one equation that it remains valid without the theta (θ) appearing in this new equation.

Worked-out problems on eliminate theta (θ) between the equations:

1. Eliminate theta between the equations:

x = a sin θ + b cos θ and y = a cos θ – b sin θ

OR,

If x = a sin θ + b cos θ and y = a cos θ –b sin θ, prove that

x2 + y2 = a2 + b2.

Solution:

We have x2 + y2 = (a sin θ + b cos θ)2 + (a cos θ – b sin θ)2

= (a2 sin2 θ + b2 cos2 θ + 2ab sin θ cos θ) + (a2 cos2 θ + b2 sin2 θ - 2ab sin θ cos θ)

= a2 sin2 θ + b2 cos2 θ + 2ab sin θ cos θ + a2 cos2 θ + b2 sin2 θ - 2ab sin θ cos θ

= a2 sin2 θ + b2 cos2 θ + a2 cos2 θ + b2 sin2 θ

= a2 sin2 θ + a2 cos2 θ + b2 sin2 θ + b2 cos2 θ

= a2 (sin2 θ + cos2 θ) + b2 (sin2 θ + cos2 θ)

= a2 (1) + b2 (1); [since, sin2 θ + cos2 θ = 1]

= a2 + b2

Therefore, x2 + y2 = a2 + b2

which is the required θ-eliminate.

2. Using the trig-identity we will solve the problems on eliminate theta (θ) between the equations:

tan θ - cot θ = a and cos θ + sin θ = b.

Solution:

tan θ – cot θ = a ………. (A)

cos θ + sin θ = b ………. (B)

Squaring both sides of (B) we get,

cos2 θ + sin2 θ + 2cos θ sin θ = b2

or, 1 + 2 cos θ sin θ = b2

or, 2 cos θ sin θ = b2 - 1 ………. (C)

Again, from (A) we get, (sin θ/cos θ) – (cos θ/sin θ) = a

or, (sin2 θ - cos2 θ)/(cos θ sin θ) = a

or, sin2θ - cos2θ = a sin θ cos θ

or, (sin θ + cos θ) (sin θ - cos θ) = a ∙ (b2 - 1)/2 ………. [by (C)]

or, b(sin θ - cos θ)= (½) a (b2 - 1) [by (B)]

or, b2 (sin θ - cos θ)2 = (1/4) a2 (b2 - 1)2, [Squaring both the sides]

or, b2 [(sin θ + cos θ)2 - 4 sinθ cos θ] = (1/4) a2 (b2 - 1)2

or, b2 [b2 - 2 ∙ (b2 - 1)] = (1/4) a2 (b2 - 1)2 [from (B) and (C)]

or, 4b2 (2 - b2) = a2 (b2 - 1)2

which is the required θ-eliminate.

Show how to use the trigonometric identities to solve the problems on eliminate theta form the given two equations.

3. x sin θ - y cos θ = √(x2 + y2) and cos2 θ/a2 + sin2 θ/b2 = 1/(x2 + y2)

Solution:

x sin θ - y cos θ = √(x2 + y2) ..........…. (A)

cos2 θ/a2 + sin2 θ/b2 = 1/(x2 + y2) ..........…. (B)

Squaring both sides of (A) we get,

x2 sin2 θ + y2 cos2 θ - 2xy sin θ cos θ = x2 + y2

or, x2 (1 - sin2 θ) + y2 (1 - cos2 θ) + 2xy sin θ cos θ = 0

or, x2 cos2 θ + y2 sin2 θ + 2 ∙ x cos θ ∙ y sin θ = 0

or, (x cos θ + y sin θ)2 = 0

or, x cos θ + y sin θ = 0

or, x cos θ = - y sin θ

or, cos θ/(-y) = sin θ/x

or, cos2 θ/y2 = sin2 θ/x2 = (cos2 θ + sin2 θ)/(y2 + x2) = 1/(x2 + y2)

Therefore, cos2 θ = y2/(x2 + y2) and sin2 θ = x2/(x2 + y2 )

Putting the values of cos2 θ and sin2 θ in (B) we get,

(1/a2) ∙ {y2/(x2} + y2) + (1/b2) ∙ {x2/(x2 + y2)} = 1/(x2 + y2)

Or, y2/a2 + x2/b2 = 1 (Since, x2 + y2 ≠0)

which is the required θ-eliminate.

The explanation will help us to understand how the steps are used technically to work-out the problems on eliminate theta form the given equations.

Trigonometric Functions