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Prove cot⁡2θ+tan⁡2θ=2sin⁡4θ\cot 2\theta + \tan 2\theta = \frac{2}{\sin 4\theta}

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Q. Prove cot⁡2θ+tan⁡2θ=2sin⁡4θ\cot 2\theta + \tan 2\theta = \frac{2}{\sin 4\theta}
  1. Write in terms of sinθ\theta and cosθ\theta: First, let's write cot2θ^2\theta and tan2θ^2\theta in terms of sinθ\theta and cosθ\theta.cot2θ=cos⁡2θsin⁡2θ\text{cot}^2\theta = \frac{\cos^2\theta}{\sin^2\theta} and tan2θ=sin⁡2θcos⁡2θ.\text{tan}^2\theta = \frac{\sin^2\theta}{\cos^2\theta}.
  2. Add expressions together: Now, let's add these two expressions together. (cot⁡2θ)+(tan⁡2θ)=cos⁡2θsin⁡2θ+sin⁡2θcos⁡2θ(\cot^2\theta) + (\tan^2\theta) = \frac{\cos^2\theta}{\sin^2\theta} + \frac{\sin^2\theta}{\cos^2\theta}.
  3. Find common denominator: To add these fractions, we need a common denominator, which is sin⁡2θcos⁡2θ\sin^2\theta\cos^2\theta. So, (cos⁡2θ)/(sin⁡2θ)+(sin⁡2θ)/(cos⁡2θ)=(cos⁡4θ+sin⁡4θ)/(sin⁡2θcos⁡2θ)(\cos^2\theta)/(\sin^2\theta) + (\sin^2\theta)/(\cos^2\theta) = (\cos^4\theta + \sin^4\theta)/(\sin^2\theta\cos^2\theta).
  4. Square sin⁡2θ+cos⁡2θ\sin^2\theta + \cos^2\theta: We know that sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1. Let's square this identity to get sin⁡4θ+2sin⁡2θcos⁡2θ+cos⁡4θ=12\sin^4\theta + 2\sin^2\theta\cos^2\theta + \cos^4\theta = 1^2.
  5. Replace sin⁡4θ+cos⁡4θ\sin^4\theta + \cos^4\theta: Now, we can replace sin⁡4θ+cos⁡4θ\sin^4\theta + \cos^4\theta in our expression with 1−2sin⁡2θcos⁡2θ1 - 2\sin^2\theta\cos^2\theta.
    (cos⁡4θ+sin⁡4θ)/(sin⁡2θcos⁡2θ)=(1−2sin⁡2θcos⁡2θ)/(sin⁡2θcos⁡2θ)(\cos^4\theta + \sin^4\theta)/(\sin^2\theta\cos^2\theta) = (1 - 2\sin^2\theta\cos^2\theta)/(\sin^2\theta\cos^2\theta).
  6. Simplify the expression: Simplify the expression by dividing both terms in the numerator by sin⁡2θcos⁡2θ\sin^2\theta\cos^2\theta.1−2sin⁡2θcos⁡2θsin⁡2θcos⁡2θ=1sin⁡2θcos⁡2θ−2\frac{1 - 2\sin^2\theta\cos^2\theta}{\sin^2\theta\cos^2\theta} = \frac{1}{\sin^2\theta\cos^2\theta} - 2.

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