9709 P3 §3.3 — Using Identities in Equations (L6)

💡 Lead In: Equations Need a Toolkit

The Scenario

Consider the equation

$$\tan\theta+\cot\theta=4, \qquad 0^{\circ}\leqslant\theta\leqslant 360^{\circ}.$$

This is not a plain $\sin\theta=k$ equation. To crack it you must first rewrite the left-hand side using an identity, then apply your equation-solving skills. Today we combine two tools: identities and solving.

Review · Activating Prior Knowledge

Quick check — pick the correct statement in each question before we solve today's equation.

Review 1 — On $[0^{\circ},360^{\circ}]$, how many solutions does $\sin\theta=0.5$ have?

$\sin\theta=0.5$ has two solutions on $[0^{\circ},360^{\circ}]$: $\theta=30^{\circ}$ (Q1) and $\theta=150^{\circ}$ (Q2). Find the reference angle, then place it in the quadrants where sine is positive (CAST rule).

Review 2 — Which pair of definitions is correct?

$\tan\theta=\dfrac{\cos\theta}{\sin\theta}, \; \cot\theta=\tan\theta$

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}, \; \cot\theta=\dfrac{1}{\tan\theta}$

$\tan\theta=\dfrac{1}{\cos\theta}, \; \cot\theta=\dfrac{1}{\sin\theta}$

$\tan\theta=\sin\theta+\cos\theta, \; \cot\theta=\sin\theta-\cos\theta$

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}$ and $\cot\theta=\dfrac{1}{\tan\theta}=\dfrac{\cos\theta}{\sin\theta}$. These let us rewrite $\tan\theta+\cot\theta$ purely in terms of $\tan\theta$.

Review 3 — In $\tan^2\theta-4\tan\theta+1=0$, what is the first move?

Treat it as a quadratic in $\tan\theta$ and solve for $\tan\theta$

Divide both sides by $\tan\theta$

Substitute $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$ immediately

Square both sides first

Write $\tan^2\theta-4\tan\theta+1=0$ as a quadratic in $\tan\theta$; solve for the two values of $\tan\theta$, then solve for $\theta$ in each case. Swapping in $\sin/\cos$ too early usually makes the algebra harder.

The Question

Rewrite $\tan\theta+\cot\theta=4$ so that it contains only $\tan\theta$, then describe the single extra step that turns it into a quadratic you can solve. Which identities and solving skills does this one equation call on?

💡 The Bridge (click to reveal)

Using $\cot\theta=\dfrac{1}{\tan\theta}$:

$$\tan\theta+\dfrac{1}{\tan\theta}=4 \;\Rightarrow\; \tan^2\theta-4\tan\theta+1=0.$$

Solving this quadratic for $\tan\theta$ and then for $\theta$ in the given range is exactly the combined skill today's lesson builds.

📖 Key Concepts: Three Strategies for Identity Equations

When an equation contains more than one trig function, the strategy is to convert everything to a single function. Three patterns appear again and again in past papers.

Strategy 1 — Convert to a single trig function:

When the equation mixes $\tan, \cot, \sec, \cosec$, write every term in $\sin$ and $\cos$, combine over a common denominator, then use identities ($\sin^2 + \cos^2 = 1$, $\sin 2\theta = 2 \sin \theta \cos \theta$) to simplify.

$$\tan \theta + \cot \theta = \dfrac{\sin \theta}{\cos \theta} + \dfrac{\cos \theta}{\sin \theta} = \dfrac{1}{\sin \theta \cos \theta} = \dfrac{2}{\sin 2\theta} = 2 \cosec 2\theta$$

Strategy 2 — Use $\sec^2 \theta = 1 + \tan^2 \theta$ to make a quadratic in $\tan$:

Equations mixing $\sec^2$ and $\tan$ can be converted to a quadratic in $\tan \theta$, which factors or uses the quadratic formula.

$$\sec^2 \theta - \tan \theta = \dfrac{5}{2} \;\Longrightarrow\; (1 + \tan^2 \theta) - \tan \theta = \dfrac{5}{2} \;\Longrightarrow\; 2\tan^2 \theta - 2\tan \theta - 3 = 0$$

Strategy 3 — Prove identities with compound + double angle formulae:

For an identity like $\cosec 2\theta + \cot 2\theta \equiv \cot \theta$, write everything in $\sin$ and $\cos$, then apply double-angle formulae to one side and simplify until both sides match.

$$\text{LHS} = \dfrac{1}{\sin 2\theta} + \dfrac{\cos 2\theta}{\sin 2\theta} = \dfrac{1 + \cos 2\theta}{\sin 2\theta} = \dfrac{2\cos^2 \theta}{2\sin \theta \cos \theta} = \dfrac{\cos \theta}{\sin \theta} = \cot \theta = \text{RHS}$$

Group · Jigsaw Activity 1: Identity Strategy Jigsaw (Jigsaw Explorer)

Work in base groups of 3. Each member becomes an expert in one identity strategy. Master your own Expert Cell, then regroup so every new group holds one expert of each type, and rebuild the solution to one equation together — no single person knows the whole path.

Round 1 — Expert Cells (5 min, silent): Each member studies only their own strategy card and taps the key identity that makes it work. You must get yours right before you can teach it to others.

Expert A · Convert to a single trig function

Write every term in $\sin$ and $\cos$, combine over a common denominator, then simplify. Which identity completes $\tan \theta + \cot \theta = \dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta} = \dfrac{1}{\sin\theta\cos\theta} = \dots$?

$\dfrac{2}{\sin 2\theta}$ $\sec 2\theta$ $\tan 2\theta$ $\dfrac{1}{\cos 2\theta}$

Expert B · Make a quadratic in tan

Mixed $\sec^2$/$\tan$ equations can be turned into a quadratic in $\tan\theta$. Which substitution makes every term depend on $\tan\theta$?

$\sec^2 \theta = 1 + \tan^2\theta$ $\sec^2 \theta = 1 - \tan^2\theta$ $\sec^2 \theta = \tan^2\theta - 1$ $\sec^2 \theta = \cosec^2\theta$

Expert C · Prove the identity

To prove an identity, convert one side to $\sin/\cos$ and apply compound + double-angle formulae. Which double-angle fact drives $\cosec 2\theta + \cot 2\theta \equiv \cot\theta$?

$\sin 2\theta = 2\sin\theta\cos\theta$ $\sin 2\theta = \sin^2\theta$ $\cos 2\theta = 2\cos^2\theta + 1$ $\tan 2\theta = \dfrac{\tan\theta}{2}$

Round 2 — Jigsaw Mix (2 min): Regroup so each new group contains exactly one Expert A, one Expert B and one Expert C. You now each hold a different piece of the puzzle.

Round 3 — Combined Jigsaw Challenge (6 min, as a group): Each expert must supply the missing step in their own box to solve $\sec^2 \theta - \tan \theta = \dfrac{5}{2}$ for $0^{\circ} \leqslant \theta \leqslant 360^{\circ}$. Tap the correct piece in your step to lock it in.

B: Replace $\sec^2\theta$ to be in terms of $t=\tan\theta$. $1 - \tan^2\theta$ $1 + \tan^2\theta$ $\tan^2\theta - 1$

→

B: Multiply through and rearrange into a quadratic in $t$. $2t^2 + 2t - 3 = 0$ $2t^2 - 2t + 3 = 0$ $2t^2 - 2t - 3 = 0$ $t^2 - t - 3 = 0$

→

A: Solve for $t = \tan\theta$. $t = \dfrac{1 \pm 3}{2}$ $t = \dfrac{1 \pm \sqrt{7}}{2}$ $t = 2 \pm \sqrt{7}$

→

C: Using the two values of $\tan\theta$, how many solutions are in $0^{\circ} \leqslant \theta \leqslant 360^{\circ}$? Two solutions No real solutions Four solutions

Individual · Predict & Verify Activity 2: Predict the Single Function (Interactive Manipulative)

An expression mixing two trig functions can often be collapsed to a single function. Here you predict the simplified form, then verify it numerically by moving a slider, then apply it to solve an equation.

Step 1 — Predict: For every value of $\theta$ (where defined), which single function equals $\tan \theta + \cot \theta$? Tap your prediction before doing any calculation. $\sin 2\theta$ $\dfrac{2}{\sin 2\theta}$ $\dfrac{1}{\cos 2\theta}$ $\cos 2\theta$

Step 2 — Verify with a slider: Drag $\theta$ and watch $\tan \theta + \cot \theta$ match your predicted single function for every value you choose.

θ: 40°

Original: tan θ + cot θ = Predicted single function:

Step 3 — Apply: Since $\tan \theta + \cot \theta = \dfrac{2}{\sin 2\theta}$, the equation $\tan \theta + \cot \theta = 4$ becomes $\dfrac{2}{\sin 2\theta} = 4 \;\Rightarrow\; \sin 2\theta = \dfrac{1}{2}$. Type the smallest positive solution $\theta$ in degrees (an integer).

θ =

Group · Guided Worked Example 1: Prove Identity Then Solve Exercise 3C Q12 9709/3 Q3 Jun 2009

(i) Prove the identity $\cosec 2\theta + \cot 2\theta \equiv \cot \theta$.

(ii) Hence solve the equation $\cosec 2\theta + \cot 2\theta = 2$, for $0^{\circ} \leqslant \theta \leqslant 360^{\circ}$.

👁 Reveal Solution

Part (i): Prove $\cosec 2\theta + \cot 2\theta \equiv \cot \theta$

Write everything in $\sin$ and $\cos$ on the LHS:

$\text{LHS} = \dfrac{1}{\sin 2\theta} + \dfrac{\cos 2\theta}{\sin 2\theta} = \dfrac{1 + \cos 2\theta}{\sin 2\theta}$

Apply double-angle formulae: $1 + \cos 2\theta = 2\cos^2 \theta$ and $\sin 2\theta = 2 \sin \theta \cos \theta$:

$= \dfrac{2\cos^2 \theta}{2 \sin \theta \cos \theta} = \dfrac{\cos \theta}{\sin \theta} = \cot \theta = \text{RHS}$ $\checkmark$

Part (ii): Solve $\cosec 2\theta + \cot 2\theta = 2$ for $0^{\circ} \leqslant \theta \leqslant 360^{\circ}$

By part (i), the equation becomes $\cot \theta = 2$, so $\tan \theta = \dfrac{1}{2}$.

Principal value: $\theta_1 = \arctan(0.5) \approx 26.565^{\circ} \approx 26.6^{\circ}$ (Q1).

Tangent is positive in Q1 and Q3, so add $180^{\circ}$:

$\theta_2 = 26.565^{\circ} + 180^{\circ} = 206.565^{\circ} \approx 206.6^{\circ}$ (Q3).

Both solutions are in $[0^{\circ}, 360^{\circ}]$.

Answers: $\boxed{\theta \approx 26.6^{\circ}, \; 206.6^{\circ}}$ (to 1 d.p.)

Individual · Guided Worked Example 2: Exact Values Without a Calculator Exercise 3C Q13 9709/32 Q3 Jun 2010

It is given that $\cos a = \dfrac{3}{5}$, where $0^{\circ} \leqslant a \leqslant 90^{\circ}$. Showing your working and without using a calculator to evaluate $a$, (i) find the exact value of $\sin(a - 30^{\circ})$, (ii) find the exact value of $\tan 2a$, and hence find the exact value of $\tan 3a$.

👁 Reveal Solution

Step 1: Find $\sin a$ and $\tan a$ from $\cos a = \tfrac{3}{5}$

$\sin^2 a = 1 - \cos^2 a = 1 - \dfrac{9}{25} = \dfrac{16}{25}$. Since $a$ is acute, $\sin a = \dfrac{4}{5}$.

$\tan a = \dfrac{\sin a}{\cos a} = \dfrac{4/5}{3/5} = \dfrac{4}{3}$.

Part (i): Find $\sin(a - 30^{\circ})$ exactly

Using $\sin(A - B) = \sin A \cos B - \cos A \sin B$ with $A = a$, $B = 30^{\circ}$, and $\cos 30^{\circ} = \dfrac{\sqrt{3}}{2}$, $\sin 30^{\circ} = \dfrac{1}{2}$:

$\sin(a - 30^{\circ}) = \sin a \cos 30^{\circ} - \cos a \sin 30^{\circ}$

$= \dfrac{4}{5} \cdot \dfrac{\sqrt{3}}{2} - \dfrac{3}{5} \cdot \dfrac{1}{2} = \dfrac{4\sqrt{3}}{10} - \dfrac{3}{10} = \boxed{\dfrac{4\sqrt{3} - 3}{10}}$

Part (ii): Find $\tan 2a$ exactly

Using $\tan 2a = \dfrac{2 \tan a}{1 - \tan^2 a}$ with $\tan a = \dfrac{4}{3}$:

$\tan 2a = \dfrac{2 \cdot \tfrac{4}{3}}{1 - \left(\tfrac{4}{3}\right)^2} = \dfrac{\tfrac{8}{3}}{1 - \tfrac{16}{9}} = \dfrac{\tfrac{8}{3}}{-\tfrac{7}{9}} = \dfrac{8}{3} \cdot \left(-\dfrac{9}{7}\right) = \boxed{-\dfrac{24}{7}}$

Part (ii) continued: Find $\tan 3a$ exactly

Using $\tan 3a = \tan(2a + a) = \dfrac{\tan 2a + \tan a}{1 - \tan 2a \cdot \tan a}$:

Substitute $\tan 2a = -\dfrac{24}{7}$ and $\tan a = \dfrac{4}{3}$:

Numerator: $-\dfrac{24}{7} + \dfrac{4}{3} = \dfrac{-72 + 28}{21} = \dfrac{-44}{21}$

Denominator: $1 - \left(-\dfrac{24}{7}\right)\left(\dfrac{4}{3}\right) = 1 + \dfrac{96}{21} = \dfrac{21 + 96}{21} = \dfrac{117}{21}$

$\tan 3a = \dfrac{-44/21}{117/21} = \dfrac{-44}{117} = \boxed{-\dfrac{44}{117}}$

✏️ Concept Quiz (Week 3 Review)

Score: 0 / 5

1. Which of the following is equivalent to $\sec x$?

$\dfrac{1}{\sin x}$

$\dfrac{1}{\cos x}$

$\dfrac{1}{\tan x}$

$\dfrac{\sin x}{\cos x}$

$\sec x = \dfrac{1}{\cos x}$. The reciprocal of $\sin$ is $\cosec$, the reciprocal of $\tan$ is $\cot$.

2. The Pythagorean identity $1 + \tan^2 x$ equals:

$\cosec^2 x$

$\cot^2 x$

$\sec^2 x$

$\cos^2 x$

$1 + \tan^2 x = \sec^2 x$. Dividing $\sin^2 x + \cos^2 x = 1$ by $\cos^2 x$ gives this identity. Similarly $1 + \cot^2 x = \cosec^2 x$.

3. The compound angle formula for $\cos(A - B)$ is:

$\cos A \cos B - \sin A \sin B$

$\cos A \cos B + \sin A \sin B$

$\sin A \cos B - \cos A \sin B$

$\sin A \cos B + \cos A \sin B$

$\cos(A - B) = \cos A \cos B + \sin A \sin B$. For cosine, the sign between the two products is opposite to the sign on the left.

4. Using a double-angle formula, $1 + \cos 2\theta$ simplifies to:

$2\sin^2 \theta$

$2\cos^2 \theta$

$2\sin \theta \cos \theta$

$1 - 2\sin^2 \theta$

Using $\cos 2\theta = 2\cos^2 \theta - 1$, we get $1 + \cos 2\theta = 2\cos^2 \theta$. (Similarly $1 - \cos 2\theta = 2\sin^2 \theta$ and $\sin 2\theta = 2 \sin \theta \cos \theta$.)

5. The expression $\tan \theta + \cot \theta$ can be rewritten (using identities) as:

$\sin \theta + \cos \theta$

$1$

$2\cosec 2\theta$

$2\tan \theta$

$\tan \theta + \cot \theta = \dfrac{\sin \theta}{\cos \theta} + \dfrac{\cos \theta}{\sin \theta} = \dfrac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \dfrac{1}{\sin \theta \cos \theta} = \dfrac{2}{\sin 2\theta} = 2\cosec 2\theta$.

🎉

You scored 0 / 4

📝 More Practice

Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.

Exercise 3C Q2 (iv), (v) Standard

Solve the following equations for $-\pi \leqslant \theta \leqslant \pi$.

(iv) $1 + \cos 2\theta = 2\sin^2 \theta$ (v) $\sin 4\theta = \cos 2\theta$

Tip (for part v): Write the expression as a $\tan$ equation in $2\theta$.

💡 Hint

For (iv): use the double-angle identity $1 + \cos 2\theta = 2\cos^2 \theta$.

For (v): write $\cos 2\theta = \sin\!\left(\dfrac{\pi}{2} - 2\theta\right)$, then solve $\sin 4\theta = \sin\!\left(\dfrac{\pi}{2} - 2\theta\right)$ using the general solution of $\sin X = \sin Y$.

👁 Reveal Solution

Part (iv): $1 + \cos 2\theta = 2\sin^2 \theta$

Use $1 + \cos 2\theta = 2\cos^2 \theta$:

$2\cos^2 \theta = 2\sin^2 \theta \;\Longrightarrow\; \cos^2 \theta - \sin^2 \theta = 0 \;\Longrightarrow\; \cos 2\theta = 0$

So $2\theta = \pm\dfrac{\pi}{2}, \pm\dfrac{3\pi}{2}$ (within $-2\pi \leqslant 2\theta \leqslant 2\pi$).

Answers: $\boxed{\theta = -\dfrac{3\pi}{4},\; -\dfrac{\pi}{4},\; \dfrac{\pi}{4},\; \dfrac{3\pi}{4}}$

Part (v): $\sin 4\theta = \cos 2\theta$

Write $\cos 2\theta = \sin\!\left(\dfrac{\pi}{2} - 2\theta\right)$, so $\sin 4\theta = \sin\!\left(\dfrac{\pi}{2} - 2\theta\right)$.

General solution of $\sin X = \sin Y$: $X = Y + 2k\pi$ or $X = \pi - Y + 2k\pi$.

Case 1: $4\theta = \dfrac{\pi}{2} - 2\theta + 2k\pi \Rightarrow 6\theta = \dfrac{\pi}{2} + 2k\pi \Rightarrow \theta = \dfrac{\pi}{12} + \dfrac{k\pi}{3}$

For $-\pi \leqslant \theta \leqslant \pi$ ($k = -3, -2, -1, 0, 1, 2$): $\theta = -\dfrac{11\pi}{12},\; -\dfrac{7\pi}{12},\; -\dfrac{\pi}{4},\; \dfrac{\pi}{12},\; \dfrac{5\pi}{12},\; \dfrac{3\pi}{4}$

Case 2: $4\theta = \pi - \left(\dfrac{\pi}{2} - 2\theta\right) + 2k\pi = \dfrac{\pi}{2} + 2\theta + 2k\pi \Rightarrow 2\theta = \dfrac{\pi}{2} + 2k\pi \Rightarrow \theta = \dfrac{\pi}{4} + k\pi$

For $-\pi \leqslant \theta \leqslant \pi$ ($k = -1, 0$): $\theta = -\dfrac{3\pi}{4},\; \dfrac{\pi}{4}$

All answers (8 solutions): $\boxed{\theta = -\dfrac{11\pi}{12},\; -\dfrac{3\pi}{4},\; -\dfrac{7\pi}{12},\; -\dfrac{\pi}{4},\; \dfrac{\pi}{12},\; \dfrac{\pi}{4},\; \dfrac{5\pi}{12},\; \dfrac{3\pi}{4}}$

Exercise 3C Q8 Medium

(i) Show that $\tan\!\left(\dfrac{\pi}{4} + \theta\right) \tan\!\left(\dfrac{\pi}{4} - \theta\right) = 1$.

(ii) Given that $\tan 26.6^{\circ} = 0.5$, solve $\tan \theta = 2$ without using your calculator. Give $\theta$ to 1 decimal place, where $0^{\circ} < \theta < 90^{\circ}$.

👁 Reveal Solution

Part (i): Show the product equals 1

Using $\tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$ with $A = \dfrac{\pi}{4}$, $B = \theta$, and $\tan \dfrac{\pi}{4} = 1$:

$\tan\!\left(\dfrac{\pi}{4} + \theta\right) = \dfrac{1 + \tan \theta}{1 - \tan \theta}$ and $\tan\!\left(\dfrac{\pi}{4} - \theta\right) = \dfrac{1 - \tan \theta}{1 + \tan \theta}$

Multiplying: $\dfrac{1 + \tan \theta}{1 - \tan \theta} \cdot \dfrac{1 - \tan \theta}{1 + \tan \theta} = 1$ $\checkmark$

Part (ii): Solve $\tan \theta = 2$ using $\tan 26.6^{\circ} = 0.5$

From part (i), $\tan\!\left(\dfrac{\pi}{4} - \alpha\right) \cdot \tan\!\left(\dfrac{\pi}{4} + \alpha\right) = 1$, so $\tan\!\left(\dfrac{\pi}{4} + \alpha\right) = \dfrac{1}{\tan\!\left(\dfrac{\pi}{4} - \alpha\right)}$.

If we set $\tan\!\left(\dfrac{\pi}{4} - \alpha\right) = 0.5 = \tan 26.6^{\circ}$, then $\dfrac{\pi}{4} - \alpha = 26.6^{\circ}$, so $\alpha = 45^{\circ} - 26.6^{\circ} = 18.4^{\circ}$.

Then $\tan\!\left(\dfrac{\pi}{4} + \alpha\right) = \dfrac{1}{0.5} = 2$, and $\dfrac{\pi}{4} + \alpha = 45^{\circ} + 18.4^{\circ} = 63.4^{\circ}$.

Answer: $\boxed{\theta = 63.4^{\circ}}$ (1 d.p.)

Exercise 3C Q9 Medium

(i) Sketch on the same axes the graphs of $y = \cos 2x$ and $y = 3\sin x - 1$ for $0 \leqslant x \leqslant 2\pi$.

(ii) Show that these curves meet at points whose $x$-coordinates are solutions of the equation $2\sin^2 x + 3\sin x - 2 = 0$.

(iii) Solve this equation to find the values of $x$ in terms of $\pi$ for $0 \leqslant x \leqslant 2\pi$.

💡 Hint

For (ii): set $\cos 2x = 3 \sin x - 1$ and use the double-angle identity $\cos 2x = 1 - 2\sin^2 x$ to eliminate $\cos 2x$. Rearrange to obtain the target quadratic.

For (iii): factorise $2\sin^2 x + 3\sin x - 2 = (2\sin x - 1)(\sin x + 2) = 0$. Note $\sin x = -2$ is impossible.

👁 Reveal Solution

Part (i): Sketch $y = \cos 2x$ and $y = 3\sin x - 1$

$y = \cos 2x$ is a cosine wave with period $\pi$ (instead of $2\pi$): it starts at $1$, hits $-1$ at $x = \dfrac{\pi}{2}$, returns to $1$ at $x = \pi$, and so on.

$y = 3\sin x - 1$ is a sine wave with amplitude $3$, shifted down by $1$: starts at $-1$, peaks at $2$ ($x = \dfrac{\pi}{2}$), back to $-1$ at $x = \pi$, trough at $-4$ ($x = \dfrac{3\pi}{2}$).

They intersect twice in $[0, 2\pi]$.

Part (ii): Show $2\sin^2 x + 3\sin x - 2 = 0$

At intersection: $\cos 2x = 3\sin x - 1$. Use $\cos 2x = 1 - 2\sin^2 x$:

$1 - 2\sin^2 x = 3\sin x - 1$

$2 - 2\sin^2 x = 3\sin x$

$2\sin^2 x + 3\sin x - 2 = 0$ $\checkmark$

Part (iii): Solve for $x$ in $[0, 2\pi]$

Factorise: $(2\sin x - 1)(\sin x + 2) = 0$, so $\sin x = \dfrac{1}{2}$ or $\sin x = -2$ (impossible).

$\sin x = \dfrac{1}{2} \Rightarrow x = \dfrac{\pi}{6}$ (Q1) or $x = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$ (Q2).

Both lie in $[0, 2\pi]$.

Answers: $\boxed{x = \dfrac{\pi}{6}, \; \dfrac{5\pi}{6}}$

Exercise 3C Q11 Challenging

Solve the equation $\tan 2x = 5 \cot x$, for $0^{\circ} < x < 180^{\circ}$.

💡 Hint 1

Let $t = \tan x$. Write $\tan 2x$ using the double-angle formula in terms of $t$, and write $\cot x = \dfrac{1}{t}$. Then equate and clear denominators.

💡 Hint 2

After substitution: $\dfrac{2t}{1 - t^2} = \dfrac{5}{t}$. Cross-multiply (noting $t \neq 0$ and $t^2 \neq 1$) to get a quadratic in $t^2$.

👁 Reveal Solution

Step 1: Substitute $t = \tan x$

$\tan 2x = \dfrac{2\tan x}{1 - \tan^2 x} = \dfrac{2t}{1 - t^2}$ and $5 \cot x = \dfrac{5}{t}$

Equation: $\dfrac{2t}{1 - t^2} = \dfrac{5}{t}$

Step 2: Clear denominators

Note $t \neq 0$ (since $\cot x$ must be defined, $x \neq 0^{\circ}, 180^{\circ}$) and $t^2 \neq 1$ (since $\tan 2x$ must be defined, $x \neq 45^{\circ}, 135^{\circ}$).

$2t \cdot t = 5(1 - t^2) \;\Longrightarrow\; 2t^2 = 5 - 5t^2 \;\Longrightarrow\; 7t^2 = 5 \;\Longrightarrow\; t^2 = \dfrac{5}{7}$

So $t = \pm\sqrt{\dfrac{5}{7}} = \pm\dfrac{\sqrt{35}}{7}$.

Step 3: Solve for $x$ in $(0^{\circ}, 180^{\circ})$

Case 1: $t = +\sqrt{\dfrac{5}{7}} \approx 0.8452$ (Q1, since $t > 0$): $x = \arctan\!\left(\sqrt{\dfrac{5}{7}}\right) \approx 40.2^{\circ}$.

Case 2: $t = -\sqrt{\dfrac{5}{7}} \approx -0.8452$ (Q2, since $0^{\circ} < x < 180^{\circ}$): $x = 180^{\circ} - 40.2^{\circ} \approx 139.8^{\circ}$.

Check: neither solution coincides with $45^{\circ}$ or $135^{\circ}$ (where $\tan 2x$ would be undefined), so both are valid.

Answers: $\boxed{x \approx 40.2^{\circ}, \; 139.8^{\circ}}$ (1 d.p.)

🎫

Exit Ticket

Before you leave, tick the objectives you feel confident about. Be honest — this guides next week's review.

I can convert an equation with multiple trig functions (e.g. $\tan \theta + \cot \theta$) into a single function using $\sin/\cos$ and double-angle identities. I can use $\sec^2 \theta = 1 + \tan^2 \theta$ to convert a mixed $\sec$/$\tan$ equation into a quadratic in $\tan \theta$. I can prove a trigonometric identity (such as $\cosec 2\theta + \cot 2\theta \equiv \cot \theta$) using compound and double-angle formulae. I can find exact values of $\sin(a \pm b)$, $\tan 2a$ and $\tan 3a$ given $\cos a$ (without evaluating $a$ on a calculator). I can solve a trig equation by rewriting it (e.g. $\tan 2x = 5 \cot x$ via $t = \tan x$) and identifying all valid solutions in a given interval.