Question

$\frac{sin2x}{1 + cos2x } = tanx$
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Answer 1

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Step-by-step explanation:

Answer 2

$\huge\underline\mathfrak\red{Answer}$

$\frac{ \sin2 \alpha }{1 + \cos2 \alpha } \\ \\ \frac{2 \sin \alpha \cos \alpha }{1 + {cos}^{2} \alpha - {sin}^{2} \alpha } \\ \\ \frac{2 \sin \alpha \cos \alpha }{(1 - {sin}^{2} \alpha ) + {cos}^{2} \alpha } \\ \\ \frac{2 \sin \alpha \cos \alpha }{ {cos}^{2} \alpha + {cos}^{2} \alpha } \\ \\ \frac{2 \sin \alpha \cos \alpha }{2 {cos}^{2} \alpha } \\ \\ \frac{ \sin \alpha }{ \cos \alpha } = \tan \alpha \\ \\ |hence \: proved|$

FORMULA USED

${sin \alpha }^{2} - {cos}^{2} \alpha = 1 \\ \\ \tan \alpha = \frac{ \sin \alpha }{ \cos \alpha }$