Prove cot2theta+tan2theta=2/(sin4theta)

Write in terms of sinθ and cosθ: First, let's write cot²θ and tan²θ in terms of sinθ and cosθ. cot²θ = (cos²θ)/(sin²θ) and tan²θ = (sin²θ)/(cos²θ).Add expressions together: Now, let's add these two expressions together. (cot²θ) + (tan²θ) = (cos²θ)/(sin²θ) + (sin²θ)/(cos²θ).Find common denominator: To add these fractions, we need a common denominator, which is sin²θcos²θ. So, (cos²θ)/(sin²θ) + (sin²θ)/(cos²θ) = (cos⁴θ + sin⁴θ)/(sin²θcos²θ).Square sin²θ + cos²θ: We know that sin²θ + cos²θ = 1. Let's square this identity to get sin⁴θ + 2sin²θcos²θ + cos⁴θ = 1².Replace sin⁴θ + cos⁴θ: Now, we can replace sin⁴θ + cos⁴θ in our expression with 1 - 2sin²θcos²θ. (cos⁴θ + sin⁴θ)/(sin²θcos²θ) = (1 - 2sin²θcos²θ)/(sin²θcos²θ).Simplify the expression: Simplify the expression by dividing both terms in the numerator by sin²θcos²θ. (1 - 2sin²θcos²θ)/(sin²θcos²θ) = 1/(sin²θcos²θ) - 2.

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

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Grade 8

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Q. Prove

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

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