$$\begin{gathered}0°\text{–}90°: Q1, \ \ 90°\text{–}180°: Q2\\[2pt] 180°\text{–}270°: Q3, \ \ 270°\text{–}360°: Q4\end{gathered}$$

$150°\to Q2$, $210°\to Q3$, $320°\to Q4$. Reflex: Start at the positive x-axis, sweep anti-clockwise, note which quarter the arm lands in.

Drop a right triangle to the x-axis: x-leg negative in Q2–3, y-leg negative in Q3–4, but the hypotenuse is always positive (it's a distance). The acute angle at the origin is the basic angle α — same fractions as before, only signs change.

Reflex: Never let the hypotenuse go negative; only the legs carry sign.

$$\tan = \frac{(-)}{(-)} > 0$$

Q1 all +, Q2 sin only, Q3 tan only, Q4 cos only. Reflex: Deduce it from the signed triangle if you ever blank on it.

$$\tan B = \frac{\sqrt{3}}{2} > 0$$

Sign clue: Q1 or Q3. Range clue ($180°<B<360°$): Q3 or Q4. Overlap → Q3. Reflex: Write both quadrant lists explicitly before drawing.

$$\begin{array}{c|ccc} & 30° & 45° & 60° \\\hline \sin & \frac{1}{2} & \frac{1}{\sqrt{2}} & \frac{\sqrt{3}}{2} \\ \cos & \frac{\sqrt{3}}{2} & \frac{1}{\sqrt{2}} & \frac{1}{2} \\ \tan & \frac{1}{\sqrt{3}} & 1 & \sqrt{3} \end{array}$$

Halved equilateral triangle ($1,\sqrt3,2$) gives 30°/60°; halved unit square ($1,1,\sqrt2$) gives 45°.

Reflex: Redraw the two triangles — thirty seconds, exact values.

$$\begin{gathered}\sin 45° = \cos 45° = \frac{1}{\sqrt{2}}\\[2pt] \tan 45° = 1\end{gathered}$$

Unit square cut on the diagonal: legs 1, 1, hypotenuse $\sqrt{2}$. Reflex: Equal legs, so sine and cosine tie and tangent is exactly 1.

$$\begin{gathered}\sec\theta = \frac{1}{\cos\theta}\\[3pt] \text{cosec}\,\theta = \frac{1}{\sin\theta}\\[3pt] \cot\theta = \frac{1}{\tan\theta}\end{gathered}$$

Find the basic ratio first, then flip it: $\frac{1}{a/b}=\frac{b}{a}$. Reflex: SEC pairs with COS, COSEC pairs with SIN — don't match by spelling.

$$\begin{gathered}\sin(90°-\theta) = \cos\theta\\[2pt] \cos(90°-\theta) = \sin\theta\\[2pt] \tan(90°-\theta) = \frac{1}{\tan\theta}\end{gathered}$$

The two acute angles in a right triangle see the same sides swapped.

Reflex: 90°−θ swaps sine ↔ cosine and flips tangent.

$$\begin{gathered}\sin(180°-\theta) = \sin\theta\\[2pt] \cos(180°+\theta) = -\cos\theta\\[2pt] \tan(360°-\theta) = -\tan\theta\\[2pt] \cos(-\theta) = \cos\theta\end{gathered}$$

Q2 only sin+, Q3 only tan+, Q4 only cos+, Q1 (+360°) unchanged. Reflex: Place the angle in its quadrant and let ASTC hand you the sign.

$$\sin(-\theta) = -\sin\theta$$

$-\theta$ sits in Q4, so sine flips — but $\cos(-\theta)=+\cos\theta$, and $\sin(a-b)\neq \sin a-\sin b$. Reflex: Trig functions aren't linear — read signs from quadrants, never factor them out.