Question 1022333
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<pre>
Suppose tan(alpha) =  {{{5/6}}} and alpha terminates in Quadrant III.
Without simplifying any square roots, calculate 


sin(alpha) =  {{{-tan(alpha)*sqrt(1/(1 + tan^2(alpha)))}}} = {{{-(5/6)*sqrt(1/(1 + (5/6)^2))}}} = {{{-(5/6)*sqrt(36/(25+36))}}} = {{{-5/sqrt(61))}}}.
               
              The sign "-" (minus) was chosen because the angle {{{alpha}}} terminates in Quadrant III.

               
cos(alpha) =  {{{-sqrt(1/(1 + tan^2(alpha)))}}} = {{{-sqrt(1/(1 + (5/6)^2))}}} = {{{-sqrt(36/(25+36))}}} = {{{-6/sqrt(61))}}}.


              The sign "-" (minus) was chosen because the angle {{{alpha}}} terminates in Quadrant III.


cot(alpha) =  {{{1/tan(alpha)}}} = {{{6/5}}}.


sec(alpha) =  {{{1/cos(alpha)}}} = {{{-sqrt(61)/6}}}.


csc(alpha) =  {{{1/sin(alpha)}}} = {{{-sqrt(61)/5}}}.
</pre>

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