Question

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

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

Answer 1

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

Given $0^{\circ }<\theta <{90}^{\circ }⟹\cos \theta >0$

$\sin \theta =\frac{8}{17}$

$\cos \theta =\sqrt{1-{\sin }^2\theta }=\sqrt{1-(\frac{8}{17}{)}^2}=\sqrt{\frac{{17}^2-8^2}{{17}^2}}=\sqrt{\frac{{15}^2}{{17}^2}}=\frac{15}{17}$

$\tan \theta +\sec \theta =\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }=\frac{1+\sin \theta }{\cos \theta }=\frac{1+\frac{8}{17}}{\frac{15}{17}}=\frac{8+17}{15}=\frac{25}{15}=\frac{5}{3}$