Trigonometry/Cosecant, Secant, Cotangent

Reciprocal identities

The cosecant ($\csc$), secant ($\sec$) and cotangent ($\cot$) functions are 'convenience' functions, just the reciprocals of (that is 1 divided by) the sine, cosine and tangent. So

$\csc (x)=\frac{1}{\sin (x)}$

$\sec (x)=\frac{1}{\cos (x)}$

$\cot (x)=\frac{1}{\tan (x)}$

Notice that cosecant is the reciprocal of sine, while from the name you might expect it to be the reciprocal of cosine!

Everything that can be done with these convenience functions can be done by writing things out in full using reciprocals of $\sin$ , $\cos$ and $\tan$ .

Unless you plan to do a great deal of trig and get familiar with working with $\csc$ , $\sec$ and $\cot$ so that you can work with them at speed, it is usually better to stay with $\sin$ , $\cos$ and $\tan$ . Just recognize these functions and be able to convert from and to them in case a question you have to answer is phrased in terms of them.

Ratio identities

Because

$\tan (x)=\frac{\sin (x)}{\cos (x)}$

and because of the definition of cotangent,

$\cot (x)=\frac{1}{\tan (x)}=\frac{1}{\frac{\sin (x)}{\cos (x)}}=\frac{\cos (x)}{\sin (x)}$

$\cot (x)=\frac{\cos (x)}{\sin (x)}$

Template:ExerciseRobox Using the definitions and what you already know about sine cos and tan:

• How close to zero can $\sec (x)$ get?
• What about $\cot (x)$ ?

Triangles for Reciprocal Trig Functions

Sine and Cosine

Tan and Sec

Cot and Cosec

The diagrams above show three triangles relating trigonometrical functions. The first one should be familiar to you from the definition of sine and cosine. The remaining two are obtained by (a) dividing all sides by $\cos (\theta )$ , and (b) dividing all sides by $\sin \theta )$ .

Because these are all right triangles we can immediately read off variants of the Pythagorean theorem, for these triangles.

${\sin }^2(\theta )+{\cos }^2(\theta )=1$

${\tan }^2(\theta )+1={\sec }^2(\theta )$

$1+{\cot }^2(\theta )={\csc }^2(\theta )$

Pythagorean Relations

The Pythagorean relations can also be derived without the diagrams.

Starting from:

${\sin }^2(a)+{\cos }^2(a)=1$

We can divide by ${\cos }^2(a)$ to get:

$\frac{{\sin }^2(a)}{{\cos }^2(a)}+1=\frac{1}{{\cos }^2(a)}$

which using the definition of tan and sec is:

${\tan }^2(a)+1={\sec }^2(a)$

---

Or again starting

${\sin }^2(a)+{\cos }^2(a)=1$

We can instead divide by ${\sin }^2(a)$ to get:

$1+\frac{{\cos }^2(a)}{{\sin }^2(a)}=\frac{1}{{\sin }^2(a)}$

which using the definitions for cot and cosec is:

$1+{\cot }^2(a)={\csc }^2(a)$

---

These formula can then be rearranged so that the 1 is on its own on one side of the equals sign, i.e from the tan relation:

${\sec }^2(a)-{\tan }^2(a)=1$

and from the cot relation

${\csc }^2(a)-{\cot }^2(a)=1$

Remembering the Formulae

It should not be necessary to remember these formulae. None of these formulae are really telling us anything new. You should be able to create them quickly from the ${\sin }^2(a)+{\cos }^2(a)=1$ relation.

Quick Check of Algebra

Template:ExampleRobox It is worth doing a quick check that the formulae are plausible. It is very easy to make a mistake with a sign. Take the last formula involving $\tan$ .

${\sec }^2(a)-{\tan }^2(a)=1$

We know that we can make a right triangle with an angle of $45^{\circ }$ and sides 1, 1 and $\sqrt{2}$ so:

$\tan (45^{\circ })=\frac{\mathrm{o}\mathrm{p}\mathrm{p}\mathrm{o}\mathrm{s}\mathrm{i}\mathrm{t}\mathrm{e}}{\mathrm{a}\mathrm{d}\mathrm{j}\mathrm{a}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}}=1$ .

$\cos (45^{\circ })=\frac{\mathrm{a}\mathrm{d}\mathrm{j}\mathrm{a}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}}{\mathrm{h}\mathrm{y}\mathrm{p}\mathrm{o}\mathrm{t}\mathrm{e}\mathrm{n}\mathrm{u}\mathrm{s}\mathrm{e}}=\frac{\sqrt{2}}{2}$

So put $a=45^{\circ }$ .

$\sec (a)=\frac{1}{\cos (45^{\circ })}=\sqrt{2}$

and

${\sec }^2(a)={\sqrt{2}}^2=2$

${\tan }^2(a)=1^2=1$

Putting those values into the equation, ${\sec }^2(a)-{\tan }^2(a)=1$ . It looks OK.

$45^{\circ }$ was probably a bad choice as it does not distinguish between $\sin$ and $\cos$ but you should get the general idea about checking the equation is plausible. Template:Robox/Close

Changed Angle Relations

Template:ExampleRobox Since $\cos (a)=\sin (90^{\circ }-a)$ , it follows (taking reciprocals of both sides) that $\sec (a)=\csc (90^{\circ }-a)$ or $\csc (a)=\sec (90^{\circ }-a)$ . Also,

$\cot (a)=\frac{\cos (a)}{\sin (a)}=\frac{\sin (90^{\circ }-a)}{\cos (90^{\circ }-a)}=\tan (90^{\circ }-a)$ .

Template:ExerciseRobox Since:

$\cos (-x)=\cos (x)$

it follows that

$\sec (-x)=\sec (x)$

If you need to, write out the missing step(s) in full.

Fill in the missing expressions on the right:

• $\csc (-x)=$

• $\sec (180^{\circ }+x)=$

• $\csc (x-180^{\circ })=$

• $\csc (90^{\circ }-\beta )=$

• $\sec (90^{\circ }-\beta )=$

• $\frac{1}{\cot (180^{\circ }-\theta )}=$

• $\sec (90^{\circ }-\alpha )=$

• $\csc (2\theta )=$

You may later want to check your answers using the graphs of the reciprocal functions that are given on a later page. Template:Robox/Close