The Scenario
On the unit circle, a point $(\cos\theta,\ \sin\theta)$ sits on a circle of radius 1. Applying Pythagoras to the right triangle between the axes and this point gives the single most-used identity in trigonometry:
$$\sin^2\theta+\cos^2\theta=1.$$
This one equation is a machine: divide it by $\cos^2\theta$ and a brand-new identity falls out. Divide it by $\sin^2\theta$ and another appears.
Review · Activating Prior Knowledge
Quick check — pick the correct statement in each question before we divide the master identity.
$\sin^2\theta+\cos^2\theta=1$ holds for every angle. Dividing it by $\cos^2\theta$ or $\sin^2\theta$ is exactly the move that produces today's new Pythagorean identities.
$\tan\theta=\dfrac{\sin\theta}{\cos\theta}$, so when we divide $\sin^2\theta+\cos^2\theta=1$ by $\cos^2\theta$, the ratio $\dfrac{\sin^2\theta}{\cos^2\theta}$ becomes $\tan^2\theta$ and $\dfrac{1}{\cos^2\theta}$ becomes $\sec^2\theta$.
$\sec\theta=\dfrac{1}{\cos\theta}$, $\cosec\theta=\dfrac{1}{\sin\theta}$ and $\cot\theta=\dfrac{1}{\tan\theta}$. These reciprocals appear when dividing the master identity: $\dfrac{1}{\cos^2\theta}=\sec^2\theta$ and $\dfrac{1}{\sin^2\theta}=\cosec^2\theta$.
The Question
Start from $\sin^2\theta+\cos^2\theta=1$. If you divide both sides by $\cos^2\theta$, which new identity appears? What if you divide by $\sin^2\theta$ instead? Before expanding each term, predict what the result will look like.
💡 The Bridge (click to reveal)
Dividing $\sin^2\theta+\cos^2\theta=1$ by $\cos^2\theta$:
$$1+\tan^2\theta=\sec^2\theta.$$
Dividing by $\sin^2\theta$:
$$\cot^2\theta+1=\cosec^2\theta.$$
These two Pythagorean-style identities are the heart of today's lesson.
Pair activity (8 min): Below are 3 proof attempts. Each contains exactly ONE error. With your partner, identify the error step — the explanation of the error is then shown directly. Score 1 point per correct identification.
Score: 0 / 3
(1 point per error found)
Individual activity (5 min): Drag each expression card into the zone for its simplified form ($1$, $\sec^2 \theta$ or $\cosec^2 \theta$). You can move cards in and out of the zones freely. When you are done, press Check to score, or Reset to start over.
Score: 0/6
☆☆☆
Solve the following equation for $0^{\circ} \leqslant x \leqslant 360^{\circ}$.
$\cosec^2 x + \cot^2 x = 2$
▼ Reveal Solution
Using $\cosec^2 x = 1 + \cot^2 x$:
$$(1 + \cot^2 x) + \cot^2 x = 2$$
$$1 + 2\cot^2 x = 2 \implies 2\cot^2 x = 1 \implies \cot^2 x = \dfrac{1}{2}$$
So $\cot x = \pm\dfrac{1}{\sqrt{2}} = \pm\dfrac{\sqrt{2}}{2}$, which means $\tan x = \pm\sqrt{2}$.
For $\tan x = \sqrt{2}$: the reference angle is $\arctan(\sqrt{2}) \approx 54.7^{\circ}$.
Since $\tan x > 0$, $x$ is in the 1st and 3rd quadrants:
$x \approx 54.7^{\circ}, \quad 180^{\circ} + 54.7^{\circ} = 234.7^{\circ}$
For $\tan x = -\sqrt{2}$: $\tan x < 0$, so $x$ is in the 2nd and 4th quadrants:
$x \approx 180^{\circ} - 54.7^{\circ} = 125.3^{\circ}, \quad 360^{\circ} - 54.7^{\circ} = 305.3^{\circ}$
$$x \approx 54.7^{\circ},\ 125.3^{\circ},\ 234.7^{\circ},\ 305.3^{\circ} \quad (\text{to 1 d.p.})$$
Solve the following equation for $0^{\circ} \leqslant x \leqslant 360^{\circ}$.
$3\sec^2 x - 10\tan x = 0$
▼ Reveal Solution
Using $\sec^2 x = 1 + \tan^2 x$:
$$3(1 + \tan^2 x) - 10\tan x = 0$$
$$3 + 3\tan^2 x - 10\tan x = 0 \implies 3\tan^2 x - 10\tan x + 3 = 0$$
Let $t = \tan x$. Factorise $3t^2 - 10t + 3 = 0$:
$$(3t - 1)(t - 3) = 0 \implies t = \dfrac{1}{3} \text{ or } t = 3$$
So $\tan x = \dfrac{1}{3}$ or $\tan x = 3$.
For $\tan x = \dfrac{1}{3}$: reference angle $\arctan\!\left(\dfrac{1}{3}\right) \approx 18.4^{\circ}$. Since $\tan x > 0$, $x$ is in the 1st and 3rd quadrants:
$x \approx 18.4^{\circ}, \quad 180^{\circ} + 18.4^{\circ} = 198.4^{\circ}$
For $\tan x = 3$: reference angle $\arctan(3) \approx 71.6^{\circ}$. Since $\tan x > 0$, $x$ is in the 1st and 3rd quadrants:
$x \approx 71.6^{\circ}, \quad 180^{\circ} + 71.6^{\circ} = 251.6^{\circ}$
$$x \approx 18.4^{\circ},\ 71.6^{\circ},\ 198.4^{\circ},\ 251.6^{\circ} \quad (\text{to 1 d.p.})$$
Test your understanding of all four lesson objectives. Each question targets one objective.
Dividing each term by $\cos^2 \theta$: $\dfrac{\sin^2 \theta}{\cos^2 \theta} + \dfrac{\cos^2 \theta}{\cos^2 \theta} = \dfrac{1}{\cos^2 \theta}$, which simplifies to $\tan^2 \theta + 1 = \sec^2 \theta$.
Dividing each term by $\sin^2 \theta$: $\dfrac{\sin^2 \theta}{\sin^2 \theta} + \dfrac{\cos^2 \theta}{\sin^2 \theta} = \dfrac{1}{\sin^2 \theta}$, which simplifies to $1 + \cot^2 \theta = \cosec^2 \theta$.
Using $\sec^2 \theta = 1 + \tan^2 \theta$: $\sec^2 \theta - \tan^2 \theta = (1 + \tan^2 \theta) - \tan^2 \theta = 1$.
Using $\sec^2 x = 1 + \tan^2 x$: $\sec^2 x = 1 + 2^2 = 1 + 4 = 5$.
Solve the following equation for $0^{\circ} \leqslant x \leqslant 360^{\circ}$.
$\cos x = \sec x$
▼ Show Solution
Since $\sec x = \dfrac{1}{\cos x}$, the equation becomes:
$$\cos x = \dfrac{1}{\cos x}$$
Multiply both sides by $\cos x$ (valid since $\sec x$ is defined only when $\cos x \neq 0$):
$$\cos^2 x = 1$$
$\cos x = 1$ or $\cos x = -1$.
For $\cos x = 1$: $x = 0^{\circ},\ 360^{\circ}$.
For $\cos x = -1$: $x = 180^{\circ}$.
At $x = 0^{\circ}, 180^{\circ}, 360^{\circ}$, $\cos x = \pm 1 \neq 0$, so $\sec x$ is defined. All solutions are valid.
$$x = 0^{\circ},\ 180^{\circ},\ 360^{\circ}$$
Solve the following equation for $0^{\circ} \leqslant x \leqslant 360^{\circ}$.
$\cosec x = \sec x$
▼ Show Solution
Since $\cosec x = \dfrac{1}{\sin x}$ and $\sec x = \dfrac{1}{\cos x}$:
$$\dfrac{1}{\sin x} = \dfrac{1}{\cos x} \implies \cos x = \sin x$$
Dividing by $\cos x$ (valid since $\sec x$ is defined, so $\cos x \neq 0$):
$$\tan x = 1$$
Reference angle: $45^{\circ}$. Since $\tan x > 0$, $x$ is in the 1st and 3rd quadrants.
$$x = 45^{\circ},\ 225^{\circ}$$
At $x = 45^{\circ}$: $\sin 45^{\circ} = \cos 45^{\circ} = \dfrac{\sqrt{2}}{2} \neq 0$, so both $\cosec x$ and $\sec x$ are defined. ✓
At $x = 225^{\circ}$: $\sin 225^{\circ} = \cos 225^{\circ} = -\dfrac{\sqrt{2}}{2} \neq 0$. ✓
$$x = 45^{\circ},\ 225^{\circ}$$
Solve the following equation for $0^{\circ} \leqslant x \leqslant 360^{\circ}$.
$2\sin x = 3\cot x$
💡 Hint
Write $\cot x = \dfrac{\cos x}{\sin x}$ and multiply through by $\sin x$ (checking $\sin x \neq 0$). Then use $\sin^2 x = 1 - \cos^2 x$ to obtain a quadratic in $\cos x$.
▼ Show Solution
Write $\cot x = \dfrac{\cos x}{\sin x}$ and multiply both sides by $\sin x$ (valid since $\cot x$ is defined, so $\sin x \neq 0$):
$$2\sin^2 x = 3\cos x$$
Substitute $\sin^2 x = 1 - \cos^2 x$:
$$2(1 - \cos^2 x) = 3\cos x \implies 2 - 2\cos^2 x = 3\cos x$$
$$2\cos^2 x + 3\cos x - 2 = 0$$
$$(2\cos x - 1)(\cos x + 2) = 0$$
So $\cos x = \dfrac{1}{2}$ or $\cos x = -2$ (rejected since $|\cos x| \leqslant 1$).
For $\cos x = \dfrac{1}{2}$: reference angle $60^{\circ}$. Since $\cos x > 0$, $x$ is in the 1st and 4th quadrants.
$$x = 60^{\circ},\ 300^{\circ}$$
Check: at $x = 60^{\circ}$, $\sin x = \dfrac{\sqrt{3}}{2} \neq 0$ ✓. At $x = 300^{\circ}$, $\sin x = -\dfrac{\sqrt{3}}{2} \neq 0$ ✓.
$$x = 60^{\circ},\ 300^{\circ}$$
Solve the following equation for $0^{\circ} \leqslant x \leqslant 360^{\circ}$.
$1 + \cot^2 x = 2\tan^2 x$
💡 Hint
Use the identity $\cosec^2 x = 1 + \cot^2 x$ to replace the left side. Then express everything in terms of $\tan x$ using $\cot^2 x = \dfrac{1}{\tan^2 x}$. Let $t = \tan^2 x$ and solve the resulting quadratic.
▼ Show Solution
Using $\cosec^2 x = 1 + \cot^2 x$, the LHS becomes $\cosec^2 x$. Alternatively, let $t = \tan^2 x$ so that $\cot^2 x = \dfrac{1}{t}$ (with $t > 0$ since $\tan x \neq 0$ for $\cot x$ to be defined):
$$1 + \dfrac{1}{t} = 2t$$
Multiply through by $t$:
$$t + 1 = 2t^2 \implies 2t^2 - t - 1 = 0$$
$$(2t + 1)(t - 1) = 0$$
So $t = -\dfrac{1}{2}$ (rejected since $t = \tan^2 x \geqslant 0$) or $t = 1$.
Since $\tan^2 x = 1$, we have $\tan x = \pm 1$.
For $\tan x = 1$: $x = 45^{\circ},\ 225^{\circ}$.
For $\tan x = -1$: $x = 135^{\circ},\ 315^{\circ}$.
At each solution, $\sin x \neq 0$ and $\cos x \neq 0$ (all are odd multiples of $45^{\circ}$), so $\tan x$ and $\cot x$ are both defined. ✓
Verify $x = 45^{\circ}$: LHS $= 1 + 1 = 2$, RHS $= 2(1) = 2$. ✓
$$x = 45^{\circ},\ 135^{\circ},\ 225^{\circ},\ 315^{\circ}$$
Exit Ticket
Check the learning objectives you feel confident about: