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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.

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

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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.

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.

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

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.

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.