Reciprocal Trigonometric Ratios

[SparlingBrandi]

~ Match each ratio in the first row with an equivalent ratio from the second row if ....
<A+<B =90degrees

<<<< 1st Row<<< Sin A Cos A Tan A Csc A Sec A Cot A >>>>>> 1st row>>>>>>
<<<< 2nd Row<<< Sin B Cos B Tan B Csc B Sec B Cot B >>>>>> 2nd Row >>>>>

I was confused as to how to approach this question

Thanks

 

[soroban]

Hello, SparlingBrandi!

Match each ratio in the first row with an equivalent ratio from the second row if \(\displaystyle \angle A + \angle B \:=\:90^o\)

. . \(\displaystyle \begin{array}{cccccc}\sin A & \cos A & \tan A & \csc A & \sec A & \cot A \\ \sin B & \cos B & \tan B & \csc B & \sec B & \cot B \end{array}\)

\(\displaystyle \text{If }\,\angle A + \angle B \:=\:90^o,\,\text{ then }A\text{ and }B\text{ are in the }same\;right\;triangle.\)

                              B
                              *
                          *   |
                 z    *       |
                  *           | y
              *               |
          *                   |
    A * - - - - - - - - - - - *
                  x

\(\displaystyle \text{We see that: }\sin A \:=\:\frac{opp}{hyp} \:=\:\frac{y}{z}\)

\(\displaystyle \text{We see that: }\cos B \:=\:\frac{adj}{hyp} \:=\:\frac{y}{x}\)

. . \(\displaystyle \text{Therefore: }\;\sin A \:=\:\cos B\)


Get the idea?

 

[fasteddie65]

Remember co-functions?
cos A = sin B
tan A = cot B
sec A = csc B
etc.