Proving an Identity with tan and cot

The reliable opening move for a trig identity is to convert everything to sines and cosines. Once $\tan$ and $\cot$ are rewritten, fractions combine and $\sin^2 + \cos^2 = 1$ does the rest. This proof shows the move in its cleanest form.

Prove the identity $\tan x + \cot x \equiv \dfrac{1}{\sin x \cos x}$. [3]

The working

Start from the left-hand side and convert to sines and cosines, using $\tan x = \frac{\sin x}{\cos x}$ and $\cot x = \frac{\cos x}{\sin x}$: $\text{LHS} = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} \quad \text{(M1)}$

Combine over the common denominator $\sin x \cos x$: $= \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} \quad \text{(M1)}$

Apply $\sin^2 x + \cos^2 x = 1$: $= \frac{1}{\sin x \cos x} = \text{RHS} \quad \blacksquare \quad \text{(A1)}$

Where the marks are won and lost

• Convert to sin/cos first. Trying to manipulate $\tan$ and $\cot$ directly rarely goes anywhere; rewriting them is the move that opens the proof.
• The common denominator produces $\sin^2 x + \cos^2 x$ in the numerator, exactly the Pythagorean identity, which collapses to $1$.
• Work down one side (the left, here) to the other. Don’t manipulate both sides.

Common mistakes

• Adding the fractions without a common denominator.
• Forgetting to apply $\sin^2 x + \cos^2 x = 1$ and leaving the numerator unsimplified.
• Writing $\cot x = \frac{\sin x}{\cos x}$ (it’s the reciprocal, $\frac{\cos x}{\sin x}$).

Full method: Trig Identities notes. See also Secant, Cosecant & Cotangent. Topic home: Trigonometry pillar.