Question: Prove that $\dfrac{\cos x}{1 - \sin x} + \dfrac{1 - \sin x}{\cos x} \equiv 2\sec x$, and hence solve $\dfrac{\cos x}{1 - \sin x} + \dfrac{1 - \sin x}{\cos x} = 5$ for $0° \leq x \leq 360°$.

Step 1. Combine LHS over a common denominator:

$$\dfrac{\cos^2 x + (1 - \sin x)^2}{(1 - \sin x)\cos x}$$

Expand numerator: $\cos^2 x + 1 - 2\sin x + \sin^2 x = 2 - 2\sin x = 2(1 - \sin x)$.

Step 2. Cancel $(1 - \sin x)$:

$$\dfrac{2(1 - \sin x)}{(1 - \sin x)\cos x} = \dfrac{2}{\cos x} = 2\sec x \;\checkmark$$

Step 3. Solve $2\sec x = 5 \Rightarrow \cos x = \dfrac{2}{5}$.

In $[0°, 360°]$: $x = \cos^{-1}\!\dfrac{2}{5} \approx 66.4°$ or $x = 360° - 66.4° = 293.6°$.

Answer: $x \approx 66.4°, 293.6°$.

Question: Prove $\dfrac{1 - \cos 2A}{1 + \cos 2A} \equiv \tan^2 A$, and hence solve $\dfrac{1 - \cos 2A}{1 + \cos 2A} = 2\tan A$ for $0 \leq A \leq 2\pi$.

Step 1. Use $1 - \cos 2A = 2\sin^2 A$ and $1 + \cos 2A = 2\cos^2 A$:

$$\dfrac{2\sin^2 A}{2\cos^2 A} = \tan^2 A \;\checkmark$$

Step 2. Replace LHS with $\tan^2 A$:

$$\tan^2 A = 2\tan A$$ $$\tan A(\tan A - 2) = 0$$ $$\tan A = 0 \quad\text{or}\quad \tan A = 2$$

Step 3. Solutions over $[0, 2\pi]$:

• $\tan A = 0$: $A = 0, \pi, 2\pi$
• $\tan A = 2$: $A = \tan^{-1} 2 \approx 1.107$ or $A \approx 1.107 + \pi \approx 4.249$

Answer: $A = 0, 1.107, \pi, 4.249, 2\pi$ rad.

Tip: When the identity is requested AND a related equation follows, the proven identity is your shortcut for the equation — substitute directly.

Question: Prove $\tan 3\theta = \dfrac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$, and hence solve $\tan 3\theta = 5\tan\theta$ for $0 < \theta < 180°$.

Step 1. Write $\tan 3\theta = \tan(2\theta + \theta)$ and use $\tan(A+B) = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}$ with $\tan 2\theta = \dfrac{2t}{1-t^2}$ where $t = \tan\theta$.

After simplification (standard derivation):

$$\tan 3\theta = \dfrac{3t - t^3}{1 - 3t^2}$$

Step 2. Set equal to $5t$:

$$\dfrac{3t - t^3}{1 - 3t^2} = 5t$$ $$3t - t^3 = 5t(1 - 3t^2) = 5t - 15t^3$$ $$14t^3 - 2t = 0 \Rightarrow 2t(7t^2 - 1) = 0$$

So $t = 0$ or $t = \pm\dfrac{1}{\sqrt{7}}$.

Step 3. In $(0°, 180°)$:

• $\tan\theta = 0 \Rightarrow$ no solution in open interval
• $\tan\theta = \dfrac{1}{\sqrt{7}} \Rightarrow \theta \approx 20.7°$
• $\tan\theta = -\dfrac{1}{\sqrt{7}} \Rightarrow \theta \approx 180° - 20.7° = 159.3°$

Answer: $\theta \approx 20.7°, 159.3°$.

Question: Solve $3\cos A - \sec A = 5\tan A$ for $0 \leq A \leq 2\pi$.

Step 1. Multiply through by $\cos A$ (note $\cos A \neq 0$):

$$3\cos^2 A - 1 = 5\sin A$$

Step 2. Use $\cos^2 A = 1 - \sin^2 A$:

$$3(1 - \sin^2 A) - 1 = 5\sin A$$ $$3 - 3\sin^2 A - 1 = 5\sin A$$ $$3\sin^2 A + 5\sin A - 2 = 0$$

Step 3. Factor:

$$(3\sin A - 1)(\sin A + 2) = 0$$ $$\sin A = \dfrac{1}{3} \quad\text{or}\quad \sin A = -2 \;(\text{rejected})$$

Step 4. $\sin A = \dfrac{1}{3}$ in $[0, 2\pi]$: $A = \sin^{-1}(1/3) \approx 0.3398$ or $A = \pi - 0.3398 \approx 2.802$.

Answer: $A \approx 0.340, 2.802$ rad.

Tip: Multiplying by $\cos A$ converts $\sec A$ and $\tan A$ into $\sin$/$\cos$ terms — clears denominators in one move.

Question: Express $9\sin\theta + 21\cos\theta$ in the form $R\sin(\theta + \alpha)$ where $R > 0$ and $0° < \alpha < 90°$. Hence solve $9\sin\theta + 21\cos\theta = 15$ for $0° \leq \theta \leq 360°$.

Step 1. $R = \sqrt{9^2 + 21^2} = \sqrt{81 + 441} = \sqrt{522} \approx 22.85$.

Step 2. $\tan\alpha = \dfrac{21}{9} = \dfrac{7}{3}$, so $\alpha = \tan^{-1}\!\dfrac{7}{3} \approx 66.8°$.

So $9\sin\theta + 21\cos\theta = \sqrt{522}\sin(\theta + 66.8°)$.

Step 3. Solve $\sqrt{522}\sin(\theta + 66.8°) = 15$:

$$\sin(\theta + 66.8°) = \dfrac{15}{\sqrt{522}} \approx 0.6565$$ $$\theta + 66.8° = 41.0° \;\text{or}\; 180° - 41.0° = 139.0°$$

(plus $360°$ multiples)

Step 4. Adjust for range:

• $\theta = 41.0° - 66.8° = -25.8°$ → add $360°$ → $334.2°$
• $\theta = 139.0° - 66.8° = 72.2°$

Answer: $\theta \approx 72.2°, 334.2°$.

Tip: Always check ALL solutions for $\theta + \alpha$, including those that put $\theta$ in range after adding $360°$.

Question: Water depth at a pier is modelled by $h = 7 + 2\sin\!\left(\dfrac{\pi t}{6}\right)$, where $h$ is in metres and $t$ is hours after midnight. Find the earliest time when $h \geq 8.6$.

Step 1. Set up the inequality:

$$7 + 2\sin\!\left(\dfrac{\pi t}{6}\right) \geq 8.6$$ $$\sin\!\left(\dfrac{\pi t}{6}\right) \geq 0.8$$

Step 2. First solution: $\dfrac{\pi t}{6} = \sin^{-1}(0.8) \approx 0.9273$ rad.

$$t = \dfrac{6 \times 0.9273}{\pi} \approx 1.77 \;\text{hours}$$

Step 3. $1.77$ hours $\approx 1$ hour $46$ minutes after midnight.

Answer: Earliest time is approximately $\mathbf{01\!:\!46}$.

Tip: For "$\geq k$" with sine, the earliest time is when sine first reaches $k$. After that, water stays above the threshold until sine drops back down — calculate exit time too if the question asks for the duration window.

Question: Sketch $y = 1 - 2\sin 3x$ for $0 \leq x \leq \pi$, and evaluate $\sin^{-1}\!\left(-\dfrac{\sqrt{3}}{2}\right)$.

Step 1. Amplitude $= 2$, period $= \dfrac{2\pi}{3}$, vertical shift $= 1$. Range $= [1-2, 1+2] = [-1, 3]$.

Sketch: starts at $(0, 1)$, dips to minimum $-1$ at $x = \dfrac{\pi}{6}$ (where $\sin 3x$ peaks at $+1$), rises to maximum $3$ at $x = \dfrac{\pi}{2}$ (where $\sin 3x = -1$), continuing the periodic pattern.

In $[0, \pi]$, the function completes $1.5$ periods.

Step 2. Evaluate $\sin^{-1}\!\left(-\dfrac{\sqrt{3}}{2}\right)$.

Principal range of $\sin^{-1}$ is $\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$. $\sin\theta = -\dfrac{\sqrt{3}}{2}$ in that range: $\theta = -\dfrac{\pi}{3}$.

Answer: $\sin^{-1}\!\left(-\dfrac{\sqrt{3}}{2}\right) = -\dfrac{\pi}{3}$.

Tip: Always sketch the curve mentally as "$1$ minus a stretched sine" — start at the vertical shift, then add/subtract the amplitude as sine swings.