Question

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

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{4}{3}$
• D. $\frac{5}{3}$

Answer 1

The correct option is D $\frac{5}{3}$

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

$\sec\theta =\dfrac { 1 }{ \cos\theta } =\dfrac { 1 }{ \sqrt { 1-{

\sin }^{ 2 }\theta } } =\dfrac { 17 }{ 15 } $

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