Explain the difference between using the trigonometric ratios (sin, cos, tan) to solve for a missing angle in a right triangle versus using the reciprocal ratios (sec, csc, cot). You must use complete sentences and any evidence needed (such as an example) to prove your point of view.
@phi @Compassionate

Question

Explain the difference between using the trigonometric ratios (sin, cos, tan) to solve for a missing angle in a right triangle versus using the reciprocal ratios (sec, csc, cot). You must use complete sentences and any evidence needed (such as an example) to prove your point of view.
@phi @Compassionate

anonymous · Answer

@Hero @e.mccormick @karatechopper  Could you help?

phi · Answer

Conceptually there is no difference.
if you learn about ratios and equations like
$$ \frac{a}{b} = \frac{c}{d} $$
that is the same thing as
$$  \frac{b}{a} = \frac{d}{c} $$
and the same is true for (example) cosine versus secant:  just "flip" both sides of the equation to switch between the two forms.
$$ \cos x = \frac{adj}{hyp} \leftrightarrow  \sec x = \frac{hyp}{adj}$$

the only reason to prefer one ratio versus the other is to avoid a divide by zero.