Question

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.

Answer 1

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.

Answer 2

tan θ – cot θ = a ………. (i)
cos θ + sin θ = b ………. (ii) 

Squaring both sides of (B) we get, 

cos2 θ + sin2 θ + 2cos θ sin θ = b2
= 1 + 2 cos θ sin θ = b2
= 2 cos θ sin θ = b2 - 1 ………. (iii) 

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

= (sin2 θ - cos2 θ)/(cos θ sin θ) = a 
= sin2θ - cos2θ = a sin θ cos θ
= (sin θ + cos θ) (sin θ - cos θ) = a ∙ (b2 - 1)/2 ………. [by (iii)]
= b(sin θ - cos θ)= (½) a (b2 - 1) [by (ii)] 
= b2 (sin θ - cos θ)2 = (1/4) a2 (b2 - 1)2, [Squaring both the sides] 
= b2 [(sin θ + cos θ)2 - 4 sinθ cos θ] = (1/4) a2 (b2 - 1)2
= b2 [b2 - 2 ∙ (b2 - 1)] = (1/4) a2 (b2 - 1)2 [from (ii) and (iii)] 
= 4b2 (2 - b2) = a2 (b2 - 1)2

which is the required θ-eliminate.