The Scenario
Yesterday you met the "building blocks" for a sum of two angles:
$$\sin(A+B)=\sin A\cos B+\cos A\sin B, \qquad \cos(A+B)=\cos A\cos B-\sin A\sin B.$$
Every trigonometric function can be written using sine and cosine. Since $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$, we can build $\tan(A+B)$ by dividing the two expansions above.
Review · Activating Prior Knowledge
Quick check — pick the correct statement in each question before we derive today's tan formula.
$\sin(A+B)=\sin A\cos B+\cos A\sin B$ and $\cos(A+B)=\cos A\cos B-\sin A\sin B$. Today we build $\tan(A+B)$ by dividing these two, so these must be rock-solid first.
$\tan\theta=\dfrac{\sin\theta}{\cos\theta}$, so $\tan(A+B)=\dfrac{\sin(A+B)}{\cos(A+B)}$. Dividing numerator and denominator by $\cos A\cos B$ will turn this into an expression in $\tan A$ and $\tan B$.
The Question
Starting from $\tan(A+B)=\dfrac{\sin(A+B)}{\cos(A+B)}$, divide numerator and denominator by $\cos A\cos B$. What do you get? Can you finish so that the answer uses only $\tan A$ and $\tan B$?
💡 The Bridge (click to reveal)
Dividing the numerator and denominator of $\dfrac{\sin(A+B)}{\cos(A+B)}$ by $\cos A\cos B$ gives:
$$\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}.$$
Replacing $B$ by $-B$ yields $\tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$. Today we apply these and use them to simplify expressions such as $\cos(x-30^{\circ})-\sqrt3\sin(x-60^{\circ})$.
Derivation: Since $\tan\theta = \dfrac{\sin\theta}{\cos\theta}$, divide the compound formula for $\sin(A+B)$ by the compound formula for $\cos(A+B)$:
$$\tan(A+B) = \dfrac{\sin(A+B)}{\cos(A+B)} = \dfrac{\sin A \cos B + \cos A \sin B}{\cos A \cos B - \sin A \sin B}$$
Dividing numerator and denominator by $\cos A \cos B$ gives the formula in terms of $\tan A$ and $\tan B$ only. The formula for $\tan(A-B)$ follows by replacing $B$ with $-B$ (and using $\tan(-B) = -\tan B$).
- Expand each term using the sin/cos compound formulae.
- Collect the $\cos x$ terms and $\sin x$ terms separately.
- Write the result as $P\cos x + Q\sin x$ where $P$ and $Q$ are constants (possibly surds).
Pair activity (8 min): A compound expression travels down an assembly line of four stations. Partner A owns Stations 1 & 3, Partner B owns Stations 2 & 4. The station owner explains their reasoning aloud, then clicks one option — one attempt per station. Keep moving until the expression is fully simplified.
Mission — express in the form $P\sin x + Q\cos x$:
$$4\sin(x + 60^{\circ}) - 2\cos(x + 30^{\circ})$$
Checkpoint discussion: Test the result at $x = 0^{\circ}$ — original: $4\sin 60^{\circ} - 2\cos 30^{\circ} = 2\sqrt{3} - \sqrt{3} = \sqrt{3}$; simplified: $3\sin 0^{\circ} + \sqrt{3}\cos 0^{\circ} = \sqrt{3}$. ✓
With your partner, discuss: does agreement at $x = 0^{\circ}$ alone prove the simplification is correct? Which other value of $x$ would you test, and why?
Individual activity (5 min): Tap an expression on the left, then tap its exact value on the right — each tap-draws a coloured line marking your match. When all 6 lines are drawn, press Check to score your work (green = correct, red = rethink). Press Reset any time to clear the board and start again. Two of the left cards need reverse recognition — spotting the $\tan(A \pm B)$ pattern hiding inside a fraction.
Tap a card on the left, then its match on the right to draw a line.
The angle $\alpha$ lies between $0^{\circ}$ and $90^{\circ}$ and is such that $2\tan^2 \alpha + \sec^2 \alpha = 5 - 4\tan \alpha$.
(i) Show that $3\tan^2 \alpha + 4\tan \alpha - 4 = 0$ and hence find the exact value of $\tan \alpha$.
(ii) It is given that the angle $\beta$ is such that $\cot(\alpha + \beta) = 6$. Without using a calculator, find the exact value of $\cot \beta$.
(9709 Paper 33 Q3 November 2005)
👁 Reveal Solution
Use $\sec^2 \alpha = 1 + \tan^2 \alpha$:
$2\tan^2 \alpha + (1 + \tan^2 \alpha) = 5 - 4\tan \alpha$
$3\tan^2 \alpha + 1 = 5 - 4\tan \alpha$
$3\tan^2 \alpha + 4\tan \alpha - 4 = 0$ $\checkmark$
$\tan \alpha = \dfrac{-4 \pm \sqrt{16 + 48}}{6} = \dfrac{-4 \pm 8}{6}$
$\tan \alpha = \dfrac{4}{6} = \dfrac{2}{3}$ or $\tan \alpha = \dfrac{-12}{6} = -2$
Since $0^{\circ} < \alpha < 90^{\circ}$, $\tan \alpha > 0$, so $\boxed{\tan \alpha = \dfrac{2}{3}}$.
$\cot(\alpha + \beta) = 6 \Rightarrow \tan(\alpha + \beta) = \dfrac{1}{6}$
Using $\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$:
$\dfrac{\frac{2}{3} + \tan \beta}{1 - \frac{2}{3}\tan \beta} = \dfrac{1}{6}$
$6\left(\dfrac{2}{3} + \tan \beta\right) = 1 - \dfrac{2}{3}\tan \beta$
$4 + 6\tan \beta = 1 - \dfrac{2}{3}\tan \beta$
$6\tan \beta + \dfrac{2}{3}\tan \beta = 1 - 4 = -3$
$\dfrac{20}{3}\tan \beta = -3$
$\tan \beta = -\dfrac{9}{20}$
$\boxed{\cot \beta = -\dfrac{20}{9}}$
(i) Show that the equation $\tan(45^{\circ} + x) - \tan x = 2$ can be written in the form $\tan^2 x + 2 \tan x - 1 = 0$.
(ii) Hence solve the equation $\tan(45^{\circ} + x) - \tan x = 2$, giving all solutions in the interval $0^{\circ} \leqslant x \leqslant 180^{\circ}$.
👁 Reveal Solution
$\tan(45^{\circ} + x) = \dfrac{\tan 45^{\circ} + \tan x}{1 - \tan 45^{\circ} \tan x} = \dfrac{1 + \tan x}{1 - \tan x}$
$\dfrac{1 + \tan x}{1 - \tan x} - \tan x = 2$
$\dfrac{1 + \tan x - \tan x(1 - \tan x)}{1 - \tan x} = 2$
$\dfrac{1 + \tan x - \tan x + \tan^2 x}{1 - \tan x} = 2$
$\dfrac{1 + \tan^2 x}{1 - \tan x} = 2$
$1 + \tan^2 x = 2(1 - \tan x) = 2 - 2\tan x$
$\tan^2 x + 2\tan x - 1 = 0$ $\checkmark$
$\tan x = \dfrac{-2 \pm \sqrt{4 + 4}}{2} = \dfrac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}$
Case 1: $\tan x = -1 + \sqrt{2} \approx 0.414$ (positive → Q1)
$x = \arctan(\sqrt{2} - 1) = \boxed{22.5^{\circ}}$
(Note: $\tan 22.5^{\circ} = \sqrt{2} - 1$ is a known exact value.)
Case 2: $\tan x = -1 - \sqrt{2} \approx -2.414$ (negative → Q2, since $0^{\circ} \leqslant x \leqslant 180^{\circ}$)
Reference angle: $\arctan(1 + \sqrt{2}) = 67.5^{\circ}$
$x = 180^{\circ} - 67.5^{\circ} = \boxed{112.5^{\circ}}$
$\boxed{x = 22.5^{\circ} \text{ or } x = 112.5^{\circ}}$
Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.
The angles $\alpha$ and $\beta$ lie in the interval $0^{\circ} < x < 180^{\circ}$, and are such that $\tan \alpha = 2 \tan \beta$ and $\tan(\alpha + \beta) = 3$. Find the possible values of $\alpha$ and $\beta$.
💡 Hint
Substitute $\tan \alpha = 2\tan \beta$ into $\tan(\alpha + \beta) = \dfrac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = 3$. This gives a quadratic in $\tan \beta$. Solve it, then find $\tan \alpha$ and the angles. Remember: both $\alpha$ and $\beta$ are in $(0^{\circ}, 180^{\circ})$, so check which quadrant each angle is in based on the sign of its tangent.
👁 Reveal Solution
$\tan(\alpha + \beta) = \dfrac{2\tan \beta + \tan \beta}{1 - 2\tan \beta \cdot \tan \beta} = \dfrac{3\tan \beta}{1 - 2\tan^2 \beta} = 3$
$3\tan \beta = 3(1 - 2\tan^2 \beta) = 3 - 6\tan^2 \beta$
$6\tan^2 \beta + 3\tan \beta - 3 = 0$
$2\tan^2 \beta + \tan \beta - 1 = 0$
$(2\tan \beta - 1)(\tan \beta + 1) = 0$
$\tan \beta = \dfrac{1}{2}$ or $\tan \beta = -1$
$\beta = \arctan\left(\dfrac{1}{2}\right) \approx 26.6^{\circ}$
$\tan \alpha = 2 \cdot \dfrac{1}{2} = 1$, so $\alpha = 45^{\circ}$
Check: $\tan(45^{\circ} + 26.6^{\circ}) = \tan(71.6^{\circ}) \approx 3$ $\checkmark$
$\beta = 180^{\circ} - 45^{\circ} = 135^{\circ}$
$\tan \alpha = 2(-1) = -2$ (negative, Q2 since $0^{\circ} < \alpha < 180^{\circ}$)
$\alpha = 180^{\circ} - \arctan(2) \approx 180^{\circ} - 63.4^{\circ} = 116.6^{\circ}$
Check: $\alpha + \beta \approx 251.6^{\circ}$, $\tan(251.6^{\circ}) = \tan(71.6^{\circ}) \approx 3$ $\checkmark$
$\boxed{\alpha = 45^{\circ},\; \beta = \arctan\left(\tfrac{1}{2}\right) \approx 26.6^{\circ}}$
or $\boxed{\alpha = 180^{\circ} - \arctan(2) \approx 116.6^{\circ},\; \beta = 135^{\circ}}$
(i) Show that the equation $\tan(30^{\circ} + \theta) = 2\tan(60^{\circ} - \theta)$ can be written in the form $\tan^2 \theta + (6\sqrt{3})\tan \theta - 5 = 0$.
(ii) Hence, or otherwise, solve the equation $\tan(30^{\circ} + \theta) = 2\tan(60^{\circ} - \theta)$, for $0^{\circ} \leqslant \theta \leqslant 180^{\circ}$.
💡 Hint
For part (i): expand both sides using $\tan(A \pm B)$. Let $t = \tan \theta$. After cross-multiplying and simplifying, you should get $t^2 + 6\sqrt{3}\,t - 5 = 0$.
For part (ii): use the quadratic formula. There are two values of $\tan \theta$ — one positive (Q1) and one negative (Q2).
👁 Reveal Solution
Let $t = \tan \theta$. Using $\tan(30^{\circ} + \theta) = \dfrac{\frac{1}{\sqrt{3}} + t}{1 - \frac{t}{\sqrt{3}}}$ and $\tan(60^{\circ} - \theta) = \dfrac{\sqrt{3} - t}{1 + \sqrt{3}\,t}$:
$\dfrac{\frac{1}{\sqrt{3}} + t}{1 - \frac{t}{\sqrt{3}}} = 2 \cdot \dfrac{\sqrt{3} - t}{1 + \sqrt{3}\,t}$
Multiply the LHS numerator and denominator by $\sqrt{3}$:
$\dfrac{1 + \sqrt{3}\,t}{\sqrt{3} - t} = \dfrac{2(\sqrt{3} - t)}{1 + \sqrt{3}\,t}$
$(1 + \sqrt{3}\,t)^2 = 2(\sqrt{3} - t)^2$
$1 + 2\sqrt{3}\,t + 3t^2 = 2(3 - 2\sqrt{3}\,t + t^2)$
$1 + 2\sqrt{3}\,t + 3t^2 = 6 - 4\sqrt{3}\,t + 2t^2$
$t^2 + 6\sqrt{3}\,t - 5 = 0$ $\checkmark$
$\tan \theta = \dfrac{-6\sqrt{3} \pm \sqrt{108 + 20}}{2} = \dfrac{-6\sqrt{3} \pm \sqrt{128}}{2} = \dfrac{-6\sqrt{3} \pm 8\sqrt{2}}{2} = -3\sqrt{3} \pm 4\sqrt{2}$
Case 1: $\tan \theta = -3\sqrt{3} + 4\sqrt{2} \approx 0.461$ (positive → Q1)
$\theta = \arctan(-3\sqrt{3} + 4\sqrt{2}) \approx \boxed{24.7^{\circ}}$
Case 2: $\tan \theta = -3\sqrt{3} - 4\sqrt{2} \approx -10.85$ (negative → Q2)
$\theta = 180^{\circ} - \arctan(3\sqrt{3} + 4\sqrt{2}) \approx 180^{\circ} - 84.7^{\circ} \approx \boxed{95.3^{\circ}}$
Solve the following equation for values of $\theta$ in the range $0^{\circ} \leqslant \theta \leqslant 180^{\circ}$.
(iv) $2\sin \theta = 3\cos(\theta - 60^{\circ})$
👁 Reveal Solution
$\cos(\theta - 60^{\circ}) = \cos \theta \cos 60^{\circ} + \sin \theta \sin 60^{\circ} = \dfrac{1}{2}\cos \theta + \dfrac{\sqrt{3}}{2}\sin \theta$
$2\sin \theta = 3\left(\dfrac{1}{2}\cos \theta + \dfrac{\sqrt{3}}{2}\sin \theta\right) = \dfrac{3}{2}\cos \theta + \dfrac{3\sqrt{3}}{2}\sin \theta$
Multiply both sides by 2:
$4\sin \theta = 3\cos \theta + 3\sqrt{3}\sin \theta$
$(4 - 3\sqrt{3})\sin \theta = 3\cos \theta$
$\tan \theta = \dfrac{3}{4 - 3\sqrt{3}}$
Rationalise: $\tan \theta = \dfrac{3(4 + 3\sqrt{3})}{(4 - 3\sqrt{3})(4 + 3\sqrt{3})} = \dfrac{3(4 + 3\sqrt{3})}{16 - 27} = \dfrac{3(4 + 3\sqrt{3})}{-11} = -\dfrac{12 + 9\sqrt{3}}{11}$
$\tan \theta \approx -2.508$ (negative → Q2, since $0^{\circ} \leqslant \theta \leqslant 180^{\circ}$)
Reference angle: $\arctan(2.508) \approx 68.3^{\circ}$
$\theta = 180^{\circ} - 68.3^{\circ} \approx \boxed{111.7^{\circ}}$ (to 1 d.p.)
The angles $A$ and $B$ are such that $\sin(A + 45^{\circ}) = (2\sqrt{2})\cos A$ and $4\sec^2 B + 5 = 12\tan B$. Without using a calculator, find the exact value of $\tan(A - B)$.
💡 Hint 1
For $\sin(A + 45^{\circ}) = (2\sqrt{2})\cos A$: expand the left side using $\sin(A + B) = \sin A \cos B + \cos A \sin B$ with $B = 45^{\circ}$. Then divide through by $\cos A$ to find $\tan A$.
💡 Hint 2
For $4\sec^2 B + 5 = 12\tan B$: use $\sec^2 B = 1 + \tan^2 B$ to get a quadratic in $\tan B$. It factorises as a perfect square.
👁 Reveal Solution
$\sin A \cos 45^{\circ} + \cos A \sin 45^{\circ} = (2\sqrt{2})\cos A$
$\dfrac{\sqrt{2}}{2}\sin A + \dfrac{\sqrt{2}}{2}\cos A = 2\sqrt{2}\cos A$
Multiply by $\dfrac{2}{\sqrt{2}}$:
$\sin A + \cos A = 4\cos A$
$\sin A = 3\cos A$
$\boxed{\tan A = 3}$
$4(1 + \tan^2 B) + 5 = 12\tan B$
$4\tan^2 B - 12\tan B + 9 = 0$
$(2\tan B - 3)^2 = 0$
$\boxed{\tan B = \dfrac{3}{2}}$
$\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B} = \dfrac{3 - \frac{3}{2}}{1 + 3 \cdot \frac{3}{2}} = \dfrac{\frac{3}{2}}{1 + \frac{9}{2}} = \dfrac{\frac{3}{2}}{\frac{11}{2}} = \boxed{\dfrac{3}{11}}$
Exit Ticket
Before you leave, tick the objectives you are confident you can achieve. Be honest — this guides next lesson's review.