general solution for tan^2 theta + cot^2 theta = 2
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Answer:
θ = 45°
Step-by-step explanation:
given, tan²θ + cot²θ = 2
=> [tex]\frac{sin^2\theta}{cos^2\theta}[/tex] + [tex]\frac{cos^2\theta}{sin^2\theta}[/tex] = 2
=> [tex]\frac{sin^4\theta+cos^4\theta}{sin^2\theta.cos^2\theta}[/tex] = 2
=> sin⁴θ +cos⁴θ = 2sin²θcos²θ ------------- (1)
we know, (sin²θ+cos²θ)² = 1²
=> sin⁴θ + 2sin²θcos²θ +cos⁴θ = 1
=> 2sin²θcos²θ = 1 - (sin⁴θ +cos⁴θ) ------------- (2)
by (1) & (2),
=> sin⁴θ +cos⁴θ = 1 - (sin⁴θ +cos⁴θ)
=> 2(sin⁴θ +cos⁴θ) = 1
=> sin⁴θ +cos⁴θ = 1/2
so (2) => 2sin²θcos²θ = 1 - (1/2)
=> 2sin²θcos²θ = 1/2
=> 4sin²θcos²θ = 1
=> 2sinθcosθ * 2sinθcosθ = 1
=> sin2θ.sin2θ = 1 (as 2sinθcosθ = sin2θ )
=> sin²2θ = 1
=> sin2θ = 1
=> 2θ = 90° (sin90° = 1)
=> θ = 90° /2
=> θ = 45°