Question

solve this friends no spam​

Answer 1

Step-by-step explanation:

$1 + {sin}^{2} \alpha = 3 \: sin \alpha \: cos \alpha \\ dividing \: whole \: equation \: by \: {cos}^{2} \alpha \\ \dfrac{1}{ {cos}^{2} \alpha } + \dfrac{ {sin}^{2} \alpha }{ {cos}^{2} \alpha } = 3 \: \dfrac{sin \alpha \: cos \alpha }{ {cos}^{2} \alpha } \\ {sec}^{2} \alpha + {tan}^{2} \alpha = 3 \: tan \alpha \\ 1 + {tan}^{2} \alpha + {tan}^{2} \alpha = 3 \: tan \alpha \\ 2 {tan}^{2} \alpha - 3 \: tan \alpha \: + 1 = 0 \\ 2 {tan}^{2} \alpha - (2 + 1)tan \alpha + 1 = 0 \\ 2 {tan}^{2} \alpha - 2tan \alpha - tan \alpha + 1 = 0 \\ 2 \: tan \alpha \: (tan \alpha - 1) - \: 1( tan \alpha - 1) = 0 \\ (tan \alpha - 1) \: (2tan \alpha - 1) = 0 \\ if \\ tan \alpha \: - 1 = 0 \\ tan \alpha = 1 \\ if \: 2tan \alpha - 1 = 0 \\ 2tan \alpha = 1 \\ tan \alpha = \dfrac{1}{2}$

Answer 2

Answer:

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