Prove: cot alpha+tan alpha=2csc 2alpha

Express in terms of sine/cosine: First, let's express cot(alpha) and tan(alpha) in terms of sine and cosine. cot(alpha) = cos(alpha)/sin(alpha) and tan(alpha) = sin(alpha)/cos(alpha).Add cot and tan: Now, let's add cot(alpha) and tan(alpha) together. cot(alpha) + tan(alpha) = (cos(alpha)/sin(alpha)) + (sin(alpha)/cos(alpha)).Find common denominator: To add these fractions, we need a common denominator, which is sin(alpha)cos(alpha). (cot(alpha) + tan(alpha)) = (cos^2(alpha) + sin^2(alpha)) / (sin(alpha)cos(alpha)).Apply Pythagorean identity: We know that cos^2(alpha) + sin^2(alpha) equals 1, according to the Pythagorean identity. So, (cot(alpha) + tan(alpha)) = 1 / (sin(alpha)cos(alpha)).Use double angle formula: Now, let's express the denominator sin(alpha)cos(alpha) in terms of the double angle formula for sine: sin(2alpha) = 2sin(alpha)cos(alpha). (cot(alpha) + tan(alpha)) = 1 / (1/2 * sin(2alpha)).Invert denominator: Inverting the fraction in the denominator, we get: (cot(alpha) + tan(alpha)) = 2 / sin(2alpha).Reciprocal of sine: Finally, we know that csc(theta) is the reciprocal of sin(theta), so csc(2alpha) = 1/sin(2alpha). Therefore, (cot(alpha) + tan(alpha)) = 2csc(2alpha).

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$1$ . Prove: $\newline$ $\cot \alpha+\tan \alpha=2 \csc 2 \alpha$

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1

. Prove:

\newline

\cot \alpha+\tan \alpha=2 \csc 2 \alpha

Express in terms of sine/cosine: First, let's express $\cot(\alpha)$ and $\tan(\alpha)$ in terms of sine and cosine. $\newline$ $\cot(\alpha) = \frac{\cos(\alpha)}{\sin(\alpha)}$ and $\tan(\alpha) = \frac{\sin(\alpha)}{\cos(\alpha)}$ .
Add cot and tan: Now, let's add $\cot(\alpha)$ and $\tan(\alpha)$ together. $\cot(\alpha) + \tan(\alpha) = \frac{\cos(\alpha)}{\sin(\alpha)} + \frac{\sin(\alpha)}{\cos(\alpha)}.$
Find common denominator: To add these fractions, we need a common denominator, which is $\sin(\alpha)\cos(\alpha)$ . $(\cot(\alpha) + \tan(\alpha)) = (\cos^2(\alpha) + \sin^2(\alpha)) / (\sin(\alpha)\cos(\alpha))$ .
Apply Pythagorean identity: We know that $\cos^2(\alpha) + \sin^2(\alpha) = 1$ , according to the Pythagorean identity. $\newline$ So, $(\cot(\alpha) + \tan(\alpha)) = \frac{1}{(\sin(\alpha)\cos(\alpha))}.$
Use double angle formula: Now, let's express the denominator $\sin(\alpha)\cos(\alpha)$ in terms of the double angle formula for sine: $\sin(2\alpha) = 2\sin(\alpha)\cos(\alpha)$ . $(\cot(\alpha) + \tan(\alpha)) = \frac{1}{\frac{1}{2} \cdot \sin(2\alpha)}$ .
Invert denominator: Inverting the fraction in the denominator, we get: $(\cot(\alpha) + \tan(\alpha)) = \frac{2}{\sin(2\alpha)}$ .
Reciprocal of sine: Finally, we know that $\csc(\theta)$ is the reciprocal of $\sin(\theta)$ , so $\csc(2\alpha) = \frac{1}{\sin(2\alpha)}$ . Therefore, $(\cot(\alpha) + \tan(\alpha)) = 2\csc(2\alpha)$ .