$$\begin{gathered}\sin^2\theta + \cos^2\theta = 1\\[3pt] 1 + \tan^2\theta = \sec^2\theta\\[3pt] 1 + \cot^2\theta = \text{cosec}^2\theta\end{gathered}$$

Divide the parent by $\cos^2\theta$ for sec, by $\sin^2\theta$ for cosec. Reflex: Re-derive the sec/cosec versions instead of misremembering them.

Start from the messier side; know your target. Convert everything to sine and cosine, combine into one fraction, factorize, cancel down to match the other side.

Reflex: Never work on both sides at once — complicated side → sin/cos → single fraction → factorize → RHS.

$$\begin{aligned}&\sin(A+B)\\ &\quad = \sin A\cos B + \cos A\sin B\\[3pt] &\cos(A+B)\\ &\quad = \cos A\cos B - \sin A\sin B\\[3pt] &\tan(A \pm B) = \tfrac{\tan A \pm \tan B}{1 \mp \tan A\tan B}\end{aligned}$$

Not algebra: $\sin(A+B) \neq \sin A + \sin B$.

Reflex: Sine keeps the sign, cosine flips it, tangent stacks tan-tan in the denominator.

$$\begin{gathered}\sin 2A = 2\sin A\cos A\\[2pt] \tan 2A = \tfrac{2\tan A}{1-\tan^2 A}\\[2pt] \cos 2A = \cos^2 A - \sin^2 A\\[2pt] = 2\cos^2 A - 1 = 1 - 2\sin^2 A\end{gathered}$$

Sin terms left over → use $1-2\sin^2A$; cos terms left over → use $2\cos^2A-1$. Reflex: Choose the cos form by what you want left over.

$$\begin{gathered}\sin A = 2\sin\tfrac{A}{2}\cos\tfrac{A}{2}\\[3pt] \cos A = 1 - 2\sin^2\tfrac{A}{2} = 2\cos^2\tfrac{A}{2} - 1\\[3pt] \tan A = \frac{2\tan(A/2)}{1-\tan^2(A/2)}\end{gathered}$$

Works for any angle and its half, not just $2x$ and $x$. Spot $A$ mixed with $\frac{A}{2}$ in a question.

Reflex: Treat 'double angle' as any angle-and-its-half pair, not just 2x and x.

$$\frac{-2\cos 2x}{\sin^2 2x}$$

LHS → M ← RHS: reduce both sides separately to this same expression, then join the chains. Reflex: Stuck? Work the other side as side working, aim for a common expression.