Recall the Trigonometric Ratio below:
[tex] \tt{ \large{sin \theta = \frac{opposite}{hypotenuse} } }\\ \tt{\large{cos \theta = \frac{adjacent}{hypotenuse} }} \\ \tt{ \large{tan \theta = \frac{opposite}{adjacent} }}[/tex]
For csc, sec and cot - they are reciprocal of sin,cos and tan.
[tex] \tt{ \large{csc \theta = \frac{1}{sin \theta} } }\\ \tt{ \large{sec \theta = \frac{1}{cos \theta} }} \\ \tt{ \large{cot \theta = \frac{1}{tan \theta} }}[/tex]
What we know now is our hypotenuse, adjacent and opposite length.
- hypotenuse = 15
- opposite = 12
- adjacent = 9
Therefore,
[tex] \large{sin \theta = \frac{12}{15} \longrightarrow \frac{4}{5} } \\ \large{cos \theta = \frac{9}{15} \longrightarrow \frac{3}{5} } \\ \large{ tan \theta = \frac{12}{9} \longrightarrow \frac{4}{3} }[/tex]As for the reciprocal of three trigonometric ratio. We just swap the numerator and denominator.
[tex] \large{csc \theta = \frac{15}{12} \longrightarrow \frac{5}{4} } \\ \large{sec \theta = \frac{15}{9} \longrightarrow \frac{5}{3} } \\ \large{cot \theta = \frac{9}{12} \longrightarrow \frac{3}{4} }[/tex]
Answer
- sin = 12/15 —> 4/5
- cos = 9/15 —> 3/5
- tan = 12/9 —> 4/3
- csc = 15/12 —> 5/4
- sec = 15/9 —> 5/3
- cot = 9/12 —> 3/4
The first is non-simplifed form while the second that has the arrow pointing is the simplest form.
Let me know if you have any doubts.
