Question: Find the exact value of $\sec\!\left(\tan^{-1}\!\dfrac{5}{2}\right)$.

Step 1. Let $\theta = \tan^{-1}\!\dfrac{5}{2}$. Then $\tan\theta = \dfrac{5}{2}$ and $\theta$ is in the principal range $-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}$.

Step 2. Construct a right triangle with opposite $= 5$, adjacent $= 2$. Hypotenuse:

$$\sqrt{5^2 + 2^2} = \sqrt{29}$$

Step 3. $\sec\theta = \dfrac{\text{hypotenuse}}{\text{adjacent}} = \dfrac{\sqrt{29}}{2}$.

Answer: $\sec\!\left(\tan^{-1}\!\dfrac{5}{2}\right) = \dfrac{\sqrt{29}}{2}$.

Tip: For inverse-trig compositions, draw a right triangle matching the given ratio. The other ratios fall out via Pythagoras.

Question: Given $\cos A = -\dfrac{3}{5}$ and $A$ is obtuse, find $\cot A$.

Step 1. Obtuse means $90° < A < 180°$, i.e. quadrant II. ASTC: in Q2, $\sin > 0$, $\cos < 0$, $\tan < 0$.

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

$$\sin^2 A = 1 - \left(-\tfrac{3}{5}\right)^2 = 1 - \tfrac{9}{25} = \tfrac{16}{25}$$ $$\sin A = \tfrac{4}{5} \;(\text{positive in Q2})$$

Step 3. $\tan A = \dfrac{\sin A}{\cos A} = \dfrac{4/5}{-3/5} = -\dfrac{4}{3}$, so:

$$\cot A = \dfrac{1}{\tan A} = -\dfrac{3}{4}$$

Tip: ASTC mnemonic — Quadrants I/II/III/IV: All / Sin / Tan / Cos are positive. Always check the sign AFTER computing the magnitude.

Question: Given $\sin A = \dfrac{3}{5}$ ($A$ acute) and $\cos B = -\dfrac{12}{13}$ ($B$ obtuse), find $\cos(A - B)$.

Step 1. Find $\cos A$ and $\sin B$.

$A$ acute → $\cos A > 0$: $\cos A = \sqrt{1 - \tfrac{9}{25}} = \dfrac{4}{5}$. $B$ obtuse → $\sin B > 0$: $\sin B = \sqrt{1 - \tfrac{144}{169}} = \dfrac{5}{13}$.

Step 2. Compound formula:

$$\cos(A - B) = \cos A \cos B + \sin A \sin B$$ $$= \left(\tfrac{4}{5}\right)\!\left(-\tfrac{12}{13}\right) + \left(\tfrac{3}{5}\right)\!\left(\tfrac{5}{13}\right)$$ $$= -\tfrac{48}{65} + \tfrac{15}{65} = -\dfrac{33}{65}$$

Tip: For compound formulas, list out all four ratios ($\sin A$, $\cos A$, $\sin B$, $\cos B$) with correct signs BEFORE substituting. Sign errors kill marks here.

Question: Given $\tan A = \dfrac{5}{12}$ and $A$ is in quadrant III, find $\tan 2A$.

Step 1. Check sign: in Q3, $\tan > 0$, so $\tan A = \dfrac{5}{12}$ (positive) is consistent. Good.

Step 2. Double-angle formula:

$$\tan 2A = \dfrac{2\tan A}{1 - \tan^2 A}$$

Step 3. Substitute $\tan A = \dfrac{5}{12}$:

$$\tan 2A = \dfrac{2 \cdot \frac{5}{12}}{1 - \left(\frac{5}{12}\right)^2} = \dfrac{\frac{10}{12}}{1 - \frac{25}{144}} = \dfrac{\frac{10}{12}}{\frac{119}{144}}$$ $$= \dfrac{10}{12} \cdot \dfrac{144}{119} = \dfrac{1440}{1428} = \dfrac{120}{119}$$

Answer: $\tan 2A = \dfrac{120}{119}$.

Tip: $\tan 2A$ is the messiest double-angle formula. Compute numerator and denominator separately, then divide. Don't try to do both at once.

Question: Given $\sin A = \dfrac{\sqrt{5}}{3}$ and $A$ is obtuse, find the exact value of $\sin\!\left(\dfrac{A}{2}\right)$.

Step 1. Find $\cos A$. $A$ obtuse → $\cos A < 0$:

$$\cos A = -\sqrt{1 - \tfrac{5}{9}} = -\sqrt{\tfrac{4}{9}} = -\dfrac{2}{3}$$

Step 2. Half-angle formula:

$$\sin^2\!\left(\tfrac{A}{2}\right) = \dfrac{1 - \cos A}{2} = \dfrac{1 - (-\tfrac{2}{3})}{2} = \dfrac{\tfrac{5}{3}}{2} = \dfrac{5}{6}$$

Step 3. Sign of $\sin(A/2)$: $A$ obtuse → $A \in (90°, 180°)$ → $A/2 \in (45°, 90°)$, which is Q1 → $\sin(A/2) > 0$.

$$\sin\!\left(\tfrac{A}{2}\right) = +\sqrt{\dfrac{5}{6}} = \dfrac{\sqrt{30}}{6}$$

⚠ Watch out: Half-angle formulas have a $\pm$ — the sign depends on which quadrant $A/2$ lands in, NOT which quadrant $A$ is in.

Question: Prove $\dfrac{\cos x + \sin x}{\cos x - \sin x} - \dfrac{\cos x - \sin x}{\cos x + \sin x} \equiv 2\tan 2x$.

Step 1. Common denominator on LHS:

$$\text{LHS} = \dfrac{(\cos x + \sin x)^2 - (\cos x - \sin x)^2}{(\cos x - \sin x)(\cos x + \sin x)}$$

Step 2. Expand numerator (difference of squares pattern):

$$(\cos x + \sin x)^2 - (\cos x - \sin x)^2 = 4\sin x \cos x = 2\sin 2x$$

Step 3. Denominator:

$$(\cos x - \sin x)(\cos x + \sin x) = \cos^2 x - \sin^2 x = \cos 2x$$

Step 4. Combine:

$$\text{LHS} = \dfrac{2\sin 2x}{\cos 2x} = 2\tan 2x = \text{RHS} \;\checkmark$$

Tip: Whenever LHS has $(a + b)^2 - (a - b)^2$ structure, the $4ab$ middle is jumping out at you. Spot the pattern and skip expanding.

Question: Solve $\cos 2x - 3\sin x - 2 = 0$ for $0 \leq x \leq 2\pi$.

Step 1. Reduce to a single function. Use $\cos 2x = 1 - 2\sin^2 x$:

$$1 - 2\sin^2 x - 3\sin x - 2 = 0$$ $$-2\sin^2 x - 3\sin x - 1 = 0$$ $$2\sin^2 x + 3\sin x + 1 = 0$$

Step 2. Factor (treat $\sin x$ as a variable):

$$(2\sin x + 1)(\sin x + 1) = 0$$ $$\sin x = -\dfrac{1}{2} \quad\text{or}\quad \sin x = -1$$

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

• $\sin x = -\tfrac{1}{2}$: $x = \pi + \tfrac{\pi}{6} = \tfrac{7\pi}{6}$ or $x = 2\pi - \tfrac{\pi}{6} = \tfrac{11\pi}{6}$
• $\sin x = -1$: $x = \tfrac{3\pi}{2}$

Answer: $x = \dfrac{7\pi}{6}, \dfrac{3\pi}{2}, \dfrac{11\pi}{6}$.

Tip: Mixed $\cos 2x$ and $\sin x$ terms — convert the $\cos 2x$ to $1 - 2\sin^2 x$ so everything is in one variable.

Question: $\triangle ABC$ is right-angled at $B$ with $AC = 12$ cm and $\angle ACB = \dfrac{\pi}{6}$. $M$ is the midpoint of $BC$. Find $\sin(\angle CAM)$ in exact form.

Step 1. Find $BC$ and $AB$. $BC = AC \cos\!\dfrac{\pi}{6} = 12 \cdot \dfrac{\sqrt{3}}{2} = 6\sqrt{3}$ $AB = AC \sin\!\dfrac{\pi}{6} = 12 \cdot \dfrac{1}{2} = 6$

Step 2. $M$ is the midpoint of $BC$: $MC = BM = 3\sqrt{3}$.

Step 3. In $\triangle ABM$ (right-angled at $B$): $AM = \sqrt{AB^2 + BM^2} = \sqrt{36 + 27} = \sqrt{63} = 3\sqrt{7}$

Step 4. In $\triangle CAM$, use sine rule or right-triangle ratios. Drop perpendicular from $A$ to $MC$: it's just $AB = 6$.

$$\sin(\angle CAM) = \dfrac{MC \cdot \sin(\angle ACM)}{AM} = \dfrac{3\sqrt{3} \cdot \tfrac{1}{2}}{3\sqrt{7}}$$

Wait — let's use a cleaner approach: $\sin(\angle CAM) = \dfrac{\text{opposite}}{\text{hypotenuse}}$ where the opposite to $\angle CAM$ in $\triangle ACM$ is... actually, by the sine rule:

$$\dfrac{MC}{\sin(\angle CAM)} = \dfrac{AM}{\sin(\angle ACM)}$$ $$\sin(\angle CAM) = \dfrac{MC \sin(\angle ACM)}{AM} = \dfrac{3\sqrt{3} \cdot \tfrac{1}{2}}{3\sqrt{7}} = \dfrac{\sqrt{3}}{2\sqrt{7}} = \dfrac{\sqrt{21}}{14}$$

Answer: $\sin(\angle CAM) = \dfrac{\sqrt{21}}{14}$.

Tip: When a midpoint is introduced, treat it as a new vertex and use the sine/cosine rule in the smaller triangle.