# 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

#### 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:

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

$\theta \quad =\quad 62$°