Question

$if \\ \: \: \: \: \sec \alpha + \tan \alpha = \sqrt{3} \\ so \: \: proved \: \: that \\ \sec\alpha = \frac{2}{ \sqrt{3} }$
​

Answer 1

Answer:

please see the attachment

Answer 2

$\huge\mathfrak\red{Answer}$

$\sec \alpha + \tan \alpha = \sqrt{3} \\ \\ \frac{1}{ \cos \alpha } + \frac{ \sin \alpha }{ \cos \alpha } = \sqrt{3} \\ \\ \frac{1 + \sin \alpha }{ \cos \alpha } = \sqrt{3} \\ \\ 1 + \sin \alpha = \sqrt{3} \cos \alpha \\ \\ squaring \\ \\ {(1 + \sin \alpha )}^{2} = {( \sqrt{3} \cos \alpha ) }^{2} \\ \\ 1 + {sin}^{2} \alpha + 2 \sin \alpha = 3 { \cos }^{2} \alpha \\ \\ 1 + {sin}^{2} \alpha + 2 \sin \alpha = 3(1 - {sin}^{2} \alpha ) \\ \\ 1 + {sin}^{2} \alpha + 2 \sin \alpha = 3 - 3 {sin}^{2} \alpha \\ \\ 1 - 3 + {sin}^{2} \alpha + 3 {sin}^{2} \alpha + 2 \sin \alpha = 0 \\ \\ - 2 + 4 { \sin }^{2} \alpha + 2 \sin \alpha = 0 \\ \\ divide \: by \: 2 \\ \\ 2 {sin}^{2} \alpha + \sin \alpha - 1 = 0 \\ \\ 2 {sin}^{2} \alpha + 2 \sin \alpha - \sin \alpha - 1 = 0 \\ \\ 2 \sin \alpha ( \sin \alpha + 1) - 1( \sin \alpha + 1) = 0 \\ \\( 2 \sin \alpha - 1)( \sin\alpha + 1) = 0 \\ \\ so \: we \: get \: \sin \alpha = \frac{1}{2} \\ \\ \alpha = \frac{\pi}{3} \\ \\ \sec \alpha = \sec \frac{\pi}{3} \\ \\ \sec \alpha = \frac{2}{ \sqrt{3} }$