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

Answer 1

Can someone help

Answer 2

@JoelSjogren

Answer 3

@crazysingh

Answer 4

I guess you will get the same answer irrespective of which method you use. I could tell better if there is any problem with such process.

Answer 5

This is the whole question

Answer 6

Okay let me think!

Answer 7

Ok thanks for helping

Answer 8

@ JoelSjogren could you help too?

Answer 9

Well, there is at least one difference. Only the set (sin, cos, tan) contains a function that is defined for all x, namely sin. This is not true for any of the functions in (sec, csc, cot). They are undefined at pi/2, 0 and 0, respectively.

Answer 10

the difference would be the orientation and location of the angle of reference

Answer 11

making use of the SOH CAH TOA
you want to first identify your angle of reference
then whatever is across that angle would be the opposite side
if you were given an angle, the opposite side, you can solve for the hypotenuse
using Sin(angle) = opposite side/h
then solve for h using algebraic manipulation

Answer 12

the adjacent side is the side next to the angle of reference. it is usually perpendicular to the opposite side

if you were given the length of the adjacent side and opposite side you can solve for the angle
you can use TOA
tangent(angle) = opposite/adjacent
then obtain the arctan of the resulting ratio

Answer 13

This implies that you cannot find the angle 0 degrees, for instance, using csc.

However, I seldom work with triangles that have 0 degree angles. In conclusion, the difference between the sets of functions is small.

Answer 14

Thanks