If $\sin { \theta } =\cfrac { 8 }{ 17 }$ where ${ 0 }^{ o }<\theta <{ 90 }^{ o }$, then $\tan { \theta } +\sec { \theta }$ is

$\sin\theta =8/17,\quad \tan\theta =\dfrac { sin\theta }{ \sqrt { 1-{ sin }^{ 2 }\theta } } =8/15$

$\sec\theta =\dfrac { 1 }{ \cos\theta } =\dfrac { 1 }{ \sqrt { 1-{ </div><div>\sin }^{ 2 }\theta } } =\dfrac { 17 }{ 15 }$

$\tan\theta +\sec\theta =\dfrac { 8 }{ 15 } +\dfrac { 17 }{ 15 } =\dfrac { 25 }{ 15 } =\dfrac { 5 }{ 3 }$