# 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: $\theta \quad =\quad { sin }^{ -1 }\frac { 15 }{ 17 }$ $\theta \quad =\quad 62$°