Question

IF $\sin \Theta =\frac{8}{17}$ where $0^{\circ }<\Theta <{90}^{\circ },$ then $\tan \Theta +sec\Theta$ is

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

Answer 1

Answer: (D)

$\frac{5}{3}$

$A{B}^2=A{C}^2-B{C}^2=225$
$\therefore AB=15$
$\Rightarrow \tan \theta =\frac{8}{15}$
and $\sec \theta =\frac{17}{15}$
$\Rightarrow \tan \theta +\sec \theta =\frac{5}{3}$