Mathematics 9709 · Pure Mathematics 2

Cosecant, secant and cotangent

$\operatorname{cosec}$ , $\sec$ and $\cot$ are just 1 over sin, cos and tan (their reciprocals).
They are 1 over something. So they go to infinity wherever the bottom ratio is zero. That one fact sets all their values and graphs.
Divide the Pythagoras identity by them and you get two new identities. These solve nearly every equation in this chapter.

Three of them: $\operatorname{cosec}\theta=\tfrac{1}{\sin\theta}$ , $\sec\theta=\tfrac{1}{\cos\theta}$ , $\cot\theta=\tfrac{1}{\tan\theta}=\tfrac{\cos\theta}{\sin\theta}$ . Watch the trap: the “co” names swap over. So $\sec$ pairs with $\cos$ , not with sin.

The three new ratios and two identities

Each new ratio is just 1 over an old one:

$\operatorname{cosec}\theta=\dfrac{1}{\sin\theta},\quad \sec\theta=\dfrac{1}{\cos\theta},\quad \cot\theta=\dfrac{1}{\tan\theta}=\dfrac{\cos\theta}{\sin\theta}$

Divide $\sin^2\theta+\cos^2\theta=1$ by $\cos^2\theta$ (then by $\sin^2\theta$ ) and two new Pythagoras identities pop out:

$1+\tan^2\theta=\sec^2\theta,\qquad 1+\cot^2\theta=\operatorname{cosec}^2\theta$

These two identities swap a squared 1-over for a plain ratio. See $\sec^2$ ? Swap it for $1+\tan^2$ . Now the equation is a quadratic in $\tan$ . Examiners test that swap over and over.

The graphs: spikes where the original is zero

cosec θ = 1/sin θ (solid) goes to infinity wherever sin θ = 0 (θ = 0°, 180°, 360°); sec θ does the same, but where cos θ = 0 (90°, 270°).

An asymptote is a line the curve gets closer and closer to but never touches. One question finds every one: where is the bottom ratio zero?
$\sec$ and $\operatorname{cosec}$ never get closer to the axis than $1$ . They stay outside the band $-1<y<1$ . Sine and cosine are the opposite: they are stuck inside it.

Worked examples

Worked example

Find the exact value of $\operatorname{cosec}240^\circ$

Fix the sign with CAST first. $240^\circ$ is in the third quadrant, where sine is negative: $\sin240^\circ=-\sin60^\circ=-\tfrac{\sqrt3}{2}$ .
Now flip it — the reciprocal of a negative is negative: $\operatorname{cosec}240^\circ=\dfrac{1}{-\sqrt3/2}=-\dfrac{2}{\sqrt3}$ .
Rationalise the surd: $-\dfrac{2}{\sqrt3}=-\dfrac{2\sqrt3}{3}$ .

Answer

\operatorname{cosec}240^\circ=-\tfrac{2\sqrt3}{3}

Worked example

Real question: show that $\sec^4\theta-\tan^4\theta\equiv1+2\tan^2\theta$

9709/31 O/N 2024 Q4(a). This “show that” uses only one identity, $\sec^2\theta=1+\tan^2\theta$ . Spot the difference of two squares first.

The left side is a difference of two squares: $\sec^4\theta-\tan^4\theta=(\sec^2\theta-\tan^2\theta)(\sec^2\theta+\tan^2\theta)$ .
The first bracket is $\sec^2\theta-\tan^2\theta=1$ (the identity, rearranged), so the product is just the second bracket: $=\sec^2\theta+\tan^2\theta$ .
Trade the leftover $\sec^2\theta$ for $1+\tan^2\theta$ : $=(1+\tan^2\theta)+\tan^2\theta=1+2\tan^2\theta$ .

Answer

\sec^4\theta-\tan^4\theta\equiv1+2\tan^2\theta\ \checkmark

Worked example

Solve $\sec^2x-\tan x=3$ for $0^\circ\le x\le360^\circ$

One squared reciprocal, so swap it for an ordinary ratio: $\sec^2x=1+\tan^2x$ , giving $1+\tan^2x-\tan x=3$ .
Tidy to a quadratic in $\tan x$ : $\tan^2x-\tan x-2=0$ , which factorises as $(\tan x-2)(\tan x+1)=0$ .
So $\tan x=2$ ( $x=63.4^\circ,\ 243.4^\circ$ ) or $\tan x=-1$ ( $x=135^\circ,\ 315^\circ$ ).

Answer

x=63.4^\circ,\ 135^\circ,\ 243.4^\circ,\ 315^\circ

Why both identities come from Pythagoras

Where 1 + tan² = sec² comes from

Take the right-angled triangle with hypotenuse $r$ and legs $x,y$ , so $x^2+y^2=r^2$ (Pythagoras). Divide everything by $x^2$ :

$1+\left(\tfrac{y}{x}\right)^2=\left(\tfrac{r}{x}\right)^2$

But $\tfrac{y}{x}=\tan\theta$ and $\tfrac{r}{x}=\sec\theta$ , so this is $1+\tan^2\theta=\sec^2\theta$ . Divide the same equation by $y^2$ instead and you get $\cot^2\theta+1=\operatorname{cosec}^2\theta$ . Both identities are just Pythagoras divided by a different leg. So you do not have to learn them as separate facts.

Where the marks go

Common mistake

Keep the sign. 1 over a negative number is still negative:

\operatorname{cosec}240^\circ

-\tfrac{2\sqrt3}{3}

, not

+\tfrac{2\sqrt3}{3}

. Use CAST to fix the sign first, then flip.

Common mistake

Watch the swap:

\sec

is 1 over

\cos

(not 1 over sin). The names not matching catches people out under pressure.

Common mistake

On a sketch, an asymptote goes where the bottom ratio is zero. Miss the one at

270^\circ

on a sec graph and you lose the sketch mark.

Now you try

Solve $2\operatorname{cosec}x-3=0$ for $0^\circ\le x\le360^\circ$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

$\operatorname{cosec}x=\tfrac32\Rightarrow \sin x=\tfrac23$ ; sine is positive in quadrants 1 and 2, so:

Answer

x=41.8^\circ\quad\text{or}\quad x=138.2^\circ

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Cosecant, secant and cotangent

$\operatorname{cosec}$ , $\sec$ and $\cot$ are just 1 over sin, cos and tan (their reciprocals).
They are 1 over something. So they go to infinity wherever the bottom ratio is zero. That one fact sets all their values and graphs.
Divide the Pythagoras identity by them and you get two new identities. These solve nearly every equation in this chapter.

The three new ratios and two identities

Each new ratio is just 1 over an old one:

$\operatorname{cosec}\theta=\dfrac{1}{\sin\theta},\quad \sec\theta=\dfrac{1}{\cos\theta},\quad \cot\theta=\dfrac{1}{\tan\theta}=\dfrac{\cos\theta}{\sin\theta}$

Divide $\sin^2\theta+\cos^2\theta=1$ by $\cos^2\theta$ (then by $\sin^2\theta$ ) and two new Pythagoras identities pop out:

$1+\tan^2\theta=\sec^2\theta,\qquad 1+\cot^2\theta=\operatorname{cosec}^2\theta$

These two identities swap a squared 1-over for a plain ratio. See $\sec^2$ ? Swap it for $1+\tan^2$ . Now the equation is a quadratic in $\tan$ . Examiners test that swap over and over.

The graphs: spikes where the original is zero

cosec θ = 1/sin θ (solid) goes to infinity wherever sin θ = 0 (θ = 0°, 180°, 360°); sec θ does the same, but where cos θ = 0 (90°, 270°).

An asymptote is a line the curve gets closer and closer to but never touches. One question finds every one: where is the bottom ratio zero?
$\sec$ and $\operatorname{cosec}$ never get closer to the axis than $1$ . They stay outside the band $-1<y<1$ . Sine and cosine are the opposite: they are stuck inside it.

Worked examples

Worked example

Find the exact value of $\operatorname{cosec}240^\circ$

Fix the sign with CAST first. $240^\circ$ is in the third quadrant, where sine is negative: $\sin240^\circ=-\sin60^\circ=-\tfrac{\sqrt3}{2}$ .
Now flip it — the reciprocal of a negative is negative: $\operatorname{cosec}240^\circ=\dfrac{1}{-\sqrt3/2}=-\dfrac{2}{\sqrt3}$ .
Rationalise the surd: $-\dfrac{2}{\sqrt3}=-\dfrac{2\sqrt3}{3}$ .

Answer

\operatorname{cosec}240^\circ=-\tfrac{2\sqrt3}{3}

Worked example

Real question: show that $\sec^4\theta-\tan^4\theta\equiv1+2\tan^2\theta$

9709/31 O/N 2024 Q4(a). This “show that” uses only one identity, $\sec^2\theta=1+\tan^2\theta$ . Spot the difference of two squares first.

The left side is a difference of two squares: $\sec^4\theta-\tan^4\theta=(\sec^2\theta-\tan^2\theta)(\sec^2\theta+\tan^2\theta)$ .
The first bracket is $\sec^2\theta-\tan^2\theta=1$ (the identity, rearranged), so the product is just the second bracket: $=\sec^2\theta+\tan^2\theta$ .
Trade the leftover $\sec^2\theta$ for $1+\tan^2\theta$ : $=(1+\tan^2\theta)+\tan^2\theta=1+2\tan^2\theta$ .

Answer

\sec^4\theta-\tan^4\theta\equiv1+2\tan^2\theta\ \checkmark

Worked example

Solve $\sec^2x-\tan x=3$ for $0^\circ\le x\le360^\circ$

One squared reciprocal, so swap it for an ordinary ratio: $\sec^2x=1+\tan^2x$ , giving $1+\tan^2x-\tan x=3$ .
Tidy to a quadratic in $\tan x$ : $\tan^2x-\tan x-2=0$ , which factorises as $(\tan x-2)(\tan x+1)=0$ .
So $\tan x=2$ ( $x=63.4^\circ,\ 243.4^\circ$ ) or $\tan x=-1$ ( $x=135^\circ,\ 315^\circ$ ).

Answer

x=63.4^\circ,\ 135^\circ,\ 243.4^\circ,\ 315^\circ

Why both identities come from Pythagoras

Where 1 + tan² = sec² comes from

Take the right-angled triangle with hypotenuse $r$ and legs $x,y$ , so $x^2+y^2=r^2$ (Pythagoras). Divide everything by $x^2$ :

$1+\left(\tfrac{y}{x}\right)^2=\left(\tfrac{r}{x}\right)^2$

Where the marks go

Common mistake

Keep the sign. 1 over a negative number is still negative:

\operatorname{cosec}240^\circ

-\tfrac{2\sqrt3}{3}

, not

+\tfrac{2\sqrt3}{3}

. Use CAST to fix the sign first, then flip.

Common mistake

Watch the swap:

\sec

is 1 over

\cos

(not 1 over sin). The names not matching catches people out under pressure.

Common mistake

On a sketch, an asymptote goes where the bottom ratio is zero. Miss the one at

270^\circ

on a sec graph and you lose the sketch mark.

Now you try

Solve $2\operatorname{cosec}x-3=0$ for $0^\circ\le x\le360^\circ$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

$\operatorname{cosec}x=\tfrac32\Rightarrow \sin x=\tfrac23$ ; sine is positive in quadrants 1 and 2, so:

Answer

x=41.8^\circ\quad\text{or}\quad x=138.2^\circ