Sec, Cosec and Cot

Summary

  • Secant (sec) is the reciprocal of cosine (cos):

sec\theta \quad =\quad \frac { hypotenuse }{ adjacent } \quad =\quad \frac { 1 }{ cos\theta }

  • Cosecant (cosec) is the reciprocal of sin:

cosec\theta \quad =\quad \frac { hypotenuse }{ opposite } \quad =\quad \frac { 1 }{ sin\theta }

  • Cotangent (cot) is the reciprocal of tan:

cot\theta \quad =\quad \frac { adjacent }{ opposite } \quad =\quad \frac { 1 }{ tan\theta }

Recall, in case of a right angle triangle if we are given one length and one angle and we have to find a missing length or if we need to find a missing angle when two lengths are given we use SOH/CAH/TOA where:

SOH is  sin\theta \quad =\quad \frac { opposite }{ hypotenuse }   shown in Fig 2

CAH is  cos\theta \quad =\quad \frac { adjacent }{ hypotenuse }   shown in Fig 3

TOA is  tan\theta \quad =\quad \frac { opposite }{ adjacent }   applies to both Fig 2 & Fig 3

However, there are more trigonometric functions i.e

cosec\theta \quad =\quad \frac { hypotenuse }{ opposite } \quad =\quad \frac { 1 }{ sin\theta }

sec\theta \quad =\quad \frac { hypotenuse }{ adjacent } \quad =\quad \frac { 1 }{ cos\theta }

cot\theta \quad =\quad \frac { adjacent }{ opposite } \quad =\quad \frac { 1 }{ tan\theta }

Note: Remember that  cosec\theta   is not the inverse of  sin\theta   and cannot be written as  { sin }^{ -1 }\theta .  This also applies to  sec\theta   and  cot\theta .

  • Secant (sec) is the reciprocal of cosine (cos)
  • Cosecant (cosec) is the reciprocal of sin
  • Cotangent (cot) is the reciprocal of tan

Graphical Comparison

 

Example 1

Q. In the triangle, find cosec⁡(A), sec⁡(A), and cot⁡(A).

Solution:

cosec\theta \quad =\quad \frac { hypotenuse }{ opposite } \quad =\quad \frac { 17 }{ 15 }

sec\theta \quad =\quad \frac { hypotenuse }{ adjacent } \quad =\quad \frac { 17 }{ 8 }

cot\theta \quad =\quad \frac { adjacent }{ opposite } \quad =\quad \frac { 8 }{ 15 }

If we need to find out the angle A, we simply choose one of the above functions i.e

cosec\theta \quad =\quad \frac { 17 }{ 15 }

We know that:

cosec\theta \quad =\quad \frac { 1 }{ sin\theta }

Hence:

\frac { 1 }{ sin\theta } \quad =\quad \frac { 17 }{ 15 }

Where:

sin\theta \quad =\quad \frac { 15 }{ 17 }

So:

\theta \quad =\quad { sin }^{ -1 }\frac { 15 }{ 17 }

\theta \quad =\quad 62°

Reference
  1. https://www.intmath.com/trigonometric-graphs/1-graphs-sine-cosine-amplitude.php
  2. https://www.intmath.com/trigonometric-graphs/4-graphs-tangent-cotangent-secant-cosecant.php
  3. http://www.nabla.hr/CL-GraphTransFun4.htm
  4. http://amsi.org.au/ESA_Senior_Years/SeniorTopic2/2d/2d_2content_6.html
  5. https://www.intmath.com/trigonometric-graphs/4-graphs-tangent-cotangent-secant-cosecant.php
  6. https://www.khanacademy.org/math/geometry-home/right-triangles-topic/reciprocal-trig-ratios-geo/a/reciprocal-trig-ratios