Prove the identity. cot 2x=(1-tan^(2)x)/(2tan x)

Express cot 2x: We start by expressing cot 2x in terms of sine and cosine functions using the identity cot(θ) = 1/tan(θ) = cos(θ)/sin(θ). cot 2x = cos(2x)/sin(2x)Use double angle formulas: Now we use the double angle formulas for sine and cosine to express cos(2x) and sin(2x) in terms of sin(x) and cos(x). cos(2x) = cos^2(x) - sin^2(x) and sin(2x) = 2sin(x)cos(x)Substitute formulas into cot 2x: Substitute the double angle formulas into the expression for cot 2x. cot 2x = (cos^2(x) - sin^2(x)) / (2sin(x)cos(x))Express sin^2(x) and cos^2(x): We know that tan(x) = sin(x)/cos(x), so we can express sin^2(x) and cos^2(x) in terms of tan(x). sin^2(x) = tan^2(x) * cos^2(x) and cos^2(x) = cos^2(x)Factor out cos^2(x): Substitute sin^2(x) and cos^2(x) in terms of tan(x) into the expression for cot 2x. cot 2x = (cos^2(x) - tan^2(x) * cos^2(x)) / (2tan(x) * cos^2(x))Cancel out cos^2(x) terms: Factor out cos^2(x) from the numerator. cot 2x = cos^2(x) * (1 - tan^2(x)) / (2tan(x) * cos^2(x))Arrive at desired identity: Cancel out the cos^2(x) terms in the numerator and denominator. cot 2x = (1 - tan^2(x)) / (2tan(x))We have arrived at the identity we were asked to prove. cot 2x = (1 - tan^2(x)) / (2tan(x))

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Prove the identity.

cot 2x=(1-tan^(2)x)/(2tan x)

$85$ . Prove the identity. $\newline$ $\cot 2 x=\frac{1-\tan ^{2} x}{2 \tan x}$

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85

. Prove the identity.

\newline

\cot 2 x=\frac{1-\tan ^{2} x}{2 \tan x}

Express $\cot 2x$ : We start by expressing $\cot 2x$ in terms of sine and cosine functions using the identity $\cot(\theta) = \frac{1}{\tan(\theta)} = \frac{\cos(\theta)}{\sin(\theta)}$ . $\newline$ $\cot 2x = \frac{\cos(2x)}{\sin(2x)}$
Use double angle formulas: Now we use the double angle formulas for sine and cosine to express $\cos(2x)$ and $\sin(2x)$ in terms of $\sin(x)$ and $\cos(x)$ . $\cos(2x) = \cos^2(x) - \sin^2(x)$ and $\sin(2x) = 2\sin(x)\cos(x)$
Substitute formulas into $\cot 2x$ : Substitute the double angle formulas into the expression for $\cot 2x$ . $\cot 2x = \frac{\cos^2(x) - \sin^2(x)}{2\sin(x)\cos(x)}$
Express $\sin^2(x)$ and $\cos^2(x)$ : We know that $\tan(x) = \frac{\sin(x)}{\cos(x)}$ , so we can express $\sin^2(x)$ and $\cos^2(x)$ in terms of $\tan(x)$ . $\newline$ $\sin^2(x) = \tan^2(x) \cdot \cos^2(x)$ and $\cos^2(x) = \cos^2(x)$
Factor out $\cos^2(x)$ : Substitute $\sin^2(x)$ and $\cos^2(x)$ in terms of $\tan(x)$ into the expression for $\cot 2x$ . $\newline$ $\cot 2x = \frac{\cos^2(x) - \tan^2(x) \cdot \cos^2(x)}{2\tan(x) \cdot \cos^2(x)}$
Cancel out $\cos^2(x)$ terms: Factor out $\cos^2(x)$ from the numerator. $\newline$ $\cot 2x = \frac{\cos^2(x) \cdot (1 - \tan^2(x))}{2\tan(x) \cdot \cos^2(x)}$
Arrive at desired identity: Cancel out the $\cos^2(x)$ terms in the numerator and denominator. $\cot 2x = \frac{1 - \tan^2(x)}{2\tan(x)}$
Arrive at desired identity: Cancel out the $\cos^2(x)$ terms in the numerator and denominator. $\cot 2x = \frac{1 - \tan^2(x)}{2\tan(x)}$ We have arrived at the identity we were asked to prove. $\cot 2x = \frac{1 - \tan^2(x)}{2\tan(x)}$