In a right angled $\displaystyle \Delta ABC$ right angled at $B$ the ratio of $AB$ to $AC$ is $\displaystyle 1:\sqrt{5}$ then $\displaystyle 3\tan \theta +5\sec ^{2}\theta$ is

$h=\sqrt { 5 } x$

$p=x$

$b=\sqrt { (\sqrt { 5 } )^ 2-x^ 2 } =\sqrt { 4{ x }^{ 2 } } =2x$

$\tan \theta =\dfrac { p }{ b } =\dfrac { 1x }{ 2x } =\dfrac { 1 }{ 2 }$

$\sec { \theta } =\dfrac { h }{ b } =\dfrac { \sqrt { 5 } x }{ 2x } =\dfrac { \sqrt { 5 } }{ 2 }$

$3\tan \theta +5\sec ^{ 2 }{ \theta =3\times } \dfrac { 1 }{ 2 } +5{ \times \left(\dfrac { \sqrt { 5 } }{ 2 } \right) }^{ 2 }=\dfrac { 3 }{ 2 } +\dfrac { 25 }{ 4 }$