Question

I went with solution​

Answer 1

$\tan( \alpha ) = \frac{3}{4} \\ on \: sqiring \: both \: side \: \\ { \tan( \alpha ) }^{2} = \frac{9}{16} \\ 1 - \ { \sec\alpha }^{2} = \frac{9}{16}$

${ \sec( \alpha ) }^{2} = 1 - \frac{9}{16} \\ { \sec( \alpha ) }^{2} = \frac{7}{16} \\ \sec( \alpha ) = \frac{ \sqrt{7} }{4}$