Proving a tan² Identity

This short proof rests on two familiar relationships: the Pythagorean identity rearranged ($1 - \cos^2 x = \sin^2 x$) and the quotient identity ($\tan x = \frac{\sin x}{\cos x}$). Spotting the first turns the whole thing into a one-line simplification.

Prove the identity $\dfrac{1 - \cos^2 x}{\cos^2 x} \equiv \tan^2 x$. [3]

The working

Start from the left-hand side and replace $1 - \cos^2 x$ using $\sin^2 x + \cos^2 x = 1$, so $1 - \cos^2 x = \sin^2 x$: $\text{LHS} = \frac{1 - \cos^2 x}{\cos^2 x} = \frac{\sin^2 x}{\cos^2 x} \quad \text{(M1, A1)}$

Now $\frac{\sin^2 x}{\cos^2 x} = \left(\frac{\sin x}{\cos x}\right)^2 = \tan^2 x$: $= \tan^2 x = \text{RHS} \quad \blacksquare \quad \text{(A1)}$

Where the marks are won and lost

• Recognising $1 - \cos^2 x = \sin^2 x$ is the decisive step, it’s the Pythagorean identity rearranged. Without it, the proof stalls.
• $\frac{\sin^2 x}{\cos^2 x} = \tan^2 x$ because $\tan x = \frac{\sin x}{\cos x}$, and squaring both sides squares the ratio.
• Work down one side (the left) to the other; don’t manipulate both.

Common mistakes

• Writing $1 - \cos^2 x = \cos^2 x$ or leaving it unsimplified.
• Forgetting that the squared ratio $\frac{\sin^2 x}{\cos^2 x}$ is $\tan^2 x$, not $\tan x$.
• Trying to cross-multiply both sides instead of simplifying one.

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