Find the Other Trig Values in Quadrant II sec(x)=-13/5

Problem

sec(x)=−13/5, Quadrant II

Solution

1. Identify the given trigonometric ratio and the quadrant. Since sec(x)=1/cos(x) we have cos(x)=−5/13 In Quadrant II, x is negative and y is positive.

2. Determine the values of r and x using the definition cos(x)=x/r Let r=13 and x=−5

3. Calculate the value of y using the Pythagorean identity x2+y2=r2

(−5)2+y2=13

25+y2=169

y2=144

y=12

Note that y is positive because the angle is in Quadrant II.

4. Find the sine and cosecant values using sin(x)=y/r and csc(x)=r/y

sin(x)=12/13

csc(x)=13/12

5. Find the tangent and cotangent values using tan(x)=y/x and cot(x)=x/y

tan(x)=−12/5

cot(x)=−5/12

Final Answer

sin(x)=12/13,cos(x)=−5/13,tan(x)=−12/5,csc(x)=13/12,cot(x)=−5/12

Want more problems? Check here!