Question

Prove that ( tan theta - cot theta )^2 isalways greater than or equal to 0

Answer 1

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Given conditions ⇒

tan θ + Cot θ = 2

We know,

Cot θ = 1/tan θ

∴ tan θ + 1/tan θ = 2

∴ (tan² θ + 1)/tan θ = 2

∴ tan²θ + 1 = 2 tanθ 

∴ tan²θ - 2 tanθ + 1 = 0

⇒ (tan θ  - 1)² = 0

⇒ tanθ - 1 = √0

⇒ tanθ - 1 = 0

∴ tanθ = 1

∴ tanθ =1

∴ tanθ = tan 45°

On Comparing,

 θ = 45°

A/c to the Question, 

 90° < θ > 0°

∴  45° lies between 90° and 0°, Hence the Value of the θ is correct.

Answer 2

tan θ + Cot θ = 2

We know,

Cot θ = 1/tan θ

∴ tan θ + 1/tan θ = 2

∴ (tan² θ + 1)/tan θ = 2

∴ tan²θ + 1 = 2 tanθ 

∴ tan²θ - 2 tanθ + 1 = 0

⇒ (tan θ  - 1)² = 0

⇒ tanθ - 1 = √0

⇒ tanθ - 1 = 0

∴ tanθ = 1

∴ tanθ =1

∴ tanθ = tan 45°

On Comparing,

 θ = 45°

A/c to the Question, 

 90° < θ > 0°

∴  45° lies between 90° and 0°, Hence the Value of the θ is correct.