Question: Prove $\dfrac{1 - \cos x}{1 + \cos x} \equiv (\csc x - \cot x)^2$.

Step 1. Start from LHS. Multiply numerator and denominator by $(1 - \cos x)$:

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

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

$$= \dfrac{(1 - \cos x)^2}{\sin^2 x} = \left(\dfrac{1 - \cos x}{\sin x}\right)^{\!2} = \left(\dfrac{1}{\sin x} - \dfrac{\cos x}{\sin x}\right)^{\!2}$$

Step 3. Recognise reciprocals:

$$= (\csc x - \cot x)^2 = \text{RHS} \;\checkmark$$

Why "multiply by the conjugate": when an identity has $1 \pm \cos x$ structure, multiplying by the conjugate creates $1 - \cos^2 x = \sin^2 x$ — the Pythagorean tool kicks in cleanly.

Question: Prove $\dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x} \equiv \sec x \csc x$.

Step 1. Combine the LHS over a common denominator:

$$\dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x} = \dfrac{\sin^2 x + \cos^2 x}{\sin x \cos x}$$

Step 2. Apply the Pythagorean identity $\sin^2 x + \cos^2 x = 1$:

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

Step 3. Split into reciprocals:

$$= \dfrac{1}{\sin x} \cdot \dfrac{1}{\cos x} = \csc x \cdot \sec x = \text{RHS} \;\checkmark$$

Tip: Whenever you see a sum of $\dfrac{p}{q} + \dfrac{q}{p}$, the common denominator gives $\dfrac{p^2 + q^2}{pq}$ — and if $p, q$ are $\sin, \cos$, the numerator becomes $1$. This is a recurring shortcut in identity proofs.

Question: Prove $\sin 2x - \cos 2x \cdot \tan x \equiv \tan x$.

Step 1. Expand $\sin 2x$ and $\cos 2x$ in $\sin x$, $\cos x$ form:

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

Step 2. Substitute into LHS:

$$\text{LHS} = 2\sin x \cos x - (2\cos^2 x - 1)\!\cdot\!\dfrac{\sin x}{\cos x}$$

Step 3. Distribute the tan term:

$$= 2\sin x \cos x - \dfrac{(2\cos^2 x - 1)\sin x}{\cos x}$$ $$= 2\sin x \cos x - 2\sin x \cos x + \dfrac{\sin x}{\cos x}$$ $$= \dfrac{\sin x}{\cos x} = \tan x = \text{RHS} \;\checkmark$$

Tip: When the identity mixes $\sin 2x$, $\cos 2x$, and $\tan x$, expand the double-angle terms in $\sin x$, $\cos x$ form first — everything becomes uniform.

Question: Show that $\cot 15° = 2 + \sqrt{3}$.

Step 1. Use $15° = 45° - 30°$. Compound formula for tangent:

$$\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B}$$ $$\tan 15° = \tan(45° - 30°) = \dfrac{1 - \tfrac{1}{\sqrt{3}}}{1 + \tfrac{1}{\sqrt{3}}}$$

Step 2. Multiply numerator and denominator by $\sqrt{3}$:

$$\tan 15° = \dfrac{\sqrt{3} - 1}{\sqrt{3} + 1}$$

Step 3. Rationalise (multiply by $\sqrt{3} - 1$):

$$\tan 15° = \dfrac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - 1^2} = \dfrac{3 - 2\sqrt{3} + 1}{2} = \dfrac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$$

Step 4. $\cot 15° = \dfrac{1}{\tan 15°} = \dfrac{1}{2 - \sqrt{3}} = \dfrac{2 + \sqrt{3}}{(2)^2 - (\sqrt{3})^2} = 2 + \sqrt{3}$ ✓

Tip: Exact-value angles like $15°$, $75°$, $105°$ are always reachable via sum/difference of $30°$, $45°$, $60°$. Rationalise at the end.

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

Step 1. Substitute $\cos 2x = 1 - 2\sin^2 x$:

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

Step 2. Factor:

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

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

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

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

Tip: The choice of which form of $\cos 2x$ to use is dictated by the OTHER terms. Here the equation has $\sin x$, so use $\cos 2x = 1 - 2\sin^2 x$ (the form that produces $\sin x$).

Question: Prove $\dfrac{1 - \cos 2x + \sin 2x}{1 + \cos 2x + \sin 2x} \equiv \tan x$.

Step 1. Use double-angle: $1 - \cos 2x = 2\sin^2 x$, $1 + \cos 2x = 2\cos^2 x$, $\sin 2x = 2\sin x \cos x$.

Step 2. Numerator:

$$1 - \cos 2x + \sin 2x = 2\sin^2 x + 2\sin x \cos x = 2\sin x(\sin x + \cos x)$$

Step 3. Denominator:

$$1 + \cos 2x + \sin 2x = 2\cos^2 x + 2\sin x \cos x = 2\cos x(\cos x + \sin x)$$

Step 4. Divide:

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

Why this works: The $1 \pm \cos 2x$ form converts cleanly into pure $\sin^2 x$ / $\cos^2 x$, which then share a common factor with $\sin 2x = 2 \sin x \cos x$. Look for this factor-and-cancel pattern.

Question: Evaluate $\sin 75° \cos 15° + \cos 75° \sin 15°$ without a calculator.

Step 1. Recognise the compound-angle pattern:

$$\sin A \cos B + \cos A \sin B = \sin(A + B)$$

with $A = 75°$, $B = 15°$.

Step 2. Apply:

$$\sin 75° \cos 15° + \cos 75° \sin 15° = \sin(75° + 15°) = \sin 90° = 1$$

Answer: $\boxed{1}$.

Tip: "Evaluate exact value" questions often have a hidden compound-angle pattern. Look for $\sin A \cos B \pm \cos A \sin B$ → $\sin(A \pm B)$, or $\cos A \cos B \pm \sin A \sin B$ → $\cos(A \mp B)$. The angles usually sum/difference to a special angle ($30°, 45°, 60°, 90°$).