Question

general solution for tan^2 theta + cot^2 theta = 2​

Answer 1

Answer:

θ = 45°

Step-by-step explanation:

given, tan²θ + cot²θ = 2

   => $\frac{sin^2\theta}{cos^2\theta}$ + $\frac{cos^2\theta}{sin^2\theta}$  = 2

   => $\frac{sin^4\theta+cos^4\theta}{sin^2\theta.cos^2\theta}$  = 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°