Tan2x Formula — 2tanx/(1−tan²x), Proof, Examples

What Is the Tan2x Formula?

The tan2x formula is the double-angle identity for the tangent function. It expresses $\tan 2x$ — the tangent of twice an angle — purely in terms of $\tan x$:

$$\tan 2x = \frac{2\tan x}{1 - \tan^2 x}.$$

Two equivalent forms are worth keeping nearby. From the ratio of the sine and cosine double-angle identities:

$$\tan 2x = \frac{\sin 2x}{\cos 2x} = \frac{2\sin x \cos x}{\cos^2 x - \sin^2 x}.$$

Note the denominator $1 - \tan^2 x$. Whenever $\tan^2 x = 1$ — that is, $\tan x = \pm 1$, which happens at $x = 45°, 135°, \dots$ — the denominator is zero and $\tan 2x$ is undefined. The formula diagnoses its own asymptotes.

How Is the Tan2x Formula Derived?

There are two clean routes, and both are worth seeing — they reinforce each other.

Method 1 — from the angle-addition identity. Write $2x = x + x$ and apply the tangent sum formula $\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A\tan B}$ with $A = B = x$:

$$\tan 2x = \tan(x + x) = \frac{\tan x + \tan x}{1 - \tan x \cdot \tan x} = \frac{2\tan x}{1 - \tan^2 x}.$$

Done in one substitution. (The tangent sum formula itself is proved in our sum and difference formulas walk-through.)

Method 2 — from $\sin 2x$ and $\cos 2x$. Start from $\tan 2x = \dfrac{\sin 2x}{\cos 2x}$, substitute the double-angle identities, then divide top and bottom by $\cos^2 x$:

$$\tan 2x = \frac{2\sin x \cos x}{\cos^2 x - \sin^2 x} = \frac{\dfrac{2\sin x \cos x}{\cos^2 x}}{\dfrac{\cos^2 x - \sin^2 x}{\cos^2 x}} = \frac{2,\frac{\sin x}{\cos x}}{1 - \frac{\sin^2 x}{\cos^2 x}} = \frac{2\tan x}{1 - \tan^2 x}.$$

Both methods land on the same identity. Method 1 is faster; Method 2 shows why the denominator is $1 - \tan^2 x$ and not something else — it falls out of dividing $\cos 2x$ by $\cos^2 x$.

Where Is the Tan2x Formula Undefined?

$\tan 2x$ is undefined in two distinct ways, and a careful student tracks both.

• Denominator zero: when $1 - \tan^2 x = 0$, i.e. $\tan x = \pm 1$. This occurs at $x = 45°, 135°, 225°, \dots$ ($x = \frac{\pi}{4} + \frac{k\pi}{2}$). Here the formula fails even though $2x$ is a perfectly ordinary angle.

• Tangent itself undefined: when $\cos 2x = 0$, i.e. $2x = 90°, 270°, \dots$, so $x = 45°, 135°, \dots$ as well. At these points $\tan 2x$ has a genuine vertical asymptote.

The period of $\tan 2x$ is $\frac{\pi}{2}$ (or $90°$) — half the period of $\tan x$, because doubling the input compresses the graph horizontally. That is the geometric meaning of the "2" inside.

Examples of the Tan2x Formula

Example 1

Given $\tan x = \dfrac{3}{4}$, find $\tan 2x$.

Substitute directly into the formula:

$$\tan 2x = \frac{2 \cdot \frac{3}{4}}{1 - \left(\frac{3}{4}\right)^2} = \frac{\frac{3}{2}}{1 - \frac{9}{16}} = \frac{\frac{3}{2}}{\frac{7}{16}} = \frac{3}{2}\cdot\frac{16}{7} = \frac{24}{7}.$$

Final answer: $\tan 2x = \dfrac{24}{7}$.

Example 2

Find $\tan 2x$ when $\sin x = \dfrac{4}{5}$ and $x$ is acute.

A tempting shortcut is to read $\tan x$ straight off the sine — to write $\tan x = \frac{4}{5}$ because the sine is $\frac{4}{5}$.

Wrong path. Taking $\tan x = \frac{4}{5}$ and pushing it through: $\tan 2x = \frac{2(4/5)}{1 - (4/5)^2} = \frac{8/5}{9/25} = \frac{40}{9}$. The answer is clean, which makes it feel right. It isn't. Tangent is sine over cosine, not sine over the hypotenuse — $\frac{4}{5}$ is $\sin x$, not $\tan x$.

The break. With $\sin x = \frac{4}{5}$ and $x$ acute, the adjacent side is $\sqrt{5^2 - 4^2} = 3$, so $\cos x = \frac{3}{5}$ and $\tan x = \frac{\sin x}{\cos x} = \frac{4}{3}$ — not $\frac{4}{5}$. The shortcut skipped the cosine entirely.

The rescue. Use $\tan x = \frac{4}{3}$:

$$\tan 2x = \frac{2 \cdot \frac{4}{3}}{1 - \left(\frac{4}{3}\right)^2} = \frac{\frac{8}{3}}{1 - \frac{16}{9}} = \frac{\frac{8}{3}}{-\frac{7}{9}} = \frac{8}{3}\cdot\left(-\frac{9}{7}\right) = -\frac{24}{7}.$$

Final answer: $\tan 2x = -\dfrac{24}{7}$. The negative sign is honest — with $\tan x = \frac{4}{3} > 1$, the angle $2x$ has crossed into a quadrant where tangent is negative.

Example 3

Find $\tan 2x$ when $x = 30°$ ($\frac{\pi}{6}$).

Here $\tan 30° = \frac{1}{\sqrt3}$, and $2x = 60°$, so we can check against the known value.

$$\tan 60° = \frac{2 \cdot \frac{1}{\sqrt3}}{1 - \frac{1}{3}} = \frac{\frac{2}{\sqrt3}}{\frac{2}{3}} = \frac{2}{\sqrt3}\cdot\frac{3}{2} = \frac{3}{\sqrt3} = \sqrt3.$$

In radians, $2x = \frac{\pi}{3}$. Final answer: $\tan 60° = \sqrt3 \approx 1.732$ — exactly the table value, so the formula checks out.

Example 4

Given $\cos x = \dfrac{12}{13}$ and $x$ acute, find $\tan 2x$.

First get $\tan x$. With $\cos x = \frac{12}{13}$, the opposite side is $\sqrt{13^2 - 12^2} = 5$, so $\sin x = \frac{5}{13}$ and $\tan x = \frac{5}{12}$.

$$\tan 2x = \frac{2 \cdot \frac{5}{12}}{1 - \left(\frac{5}{12}\right)^2} = \frac{\frac{5}{6}}{1 - \frac{25}{144}} = \frac{\frac{5}{6}}{\frac{119}{144}} = \frac{5}{6}\cdot\frac{144}{119} = \frac{120}{119}.$$

Final answer: $\tan 2x = \dfrac{120}{119} \approx 1.008$.

Example 5

Show why $\tan 2x$ is undefined at $x = 45°$ ($\frac{\pi}{4}$).

At $x = 45°$, $\tan 45° = 1$, so the denominator is $1 - 1^2 = 0$:

$$\tan(2\cdot 45°) = \frac{2(1)}{1 - 1} = \frac{2}{0} ;\rightarrow; \text{undefined}.$$

This matches the direct reading: $2x = 90°$, and $\tan 90°$ is undefined because $\cos 90° = 0$. Final answer: undefined — a vertical asymptote of $\tan 2x$ sits at $x = 45°$.

Example 6

Express $\tan 2x$ purely in terms of $\sin x$, given $\sin x = s$ and $x$ acute.

With $x$ acute, $\cos x = \sqrt{1 - s^2}$, so $\tan x = \dfrac{s}{\sqrt{1 - s^2}}$. Substitute:

$$\tan 2x = \frac{2\cdot\frac{s}{\sqrt{1-s^2}}}{1 - \frac{s^2}{1 - s^2}} = \frac{\frac{2s}{\sqrt{1-s^2}}}{\frac{1 - 2s^2}{1 - s^2}} = \frac{2s\sqrt{1 - s^2}}{1 - 2s^2}.$$

Final answer: $\tan 2x = \dfrac{2s\sqrt{1 - s^2}}{1 - 2s^2}$ — undefined when $s^2 = \frac12$, i.e. $\sin x = \frac{1}{\sqrt2}$ ($x = 45°$), exactly as Example 5 predicted.

Where the Tan2x Formula Carries Real Weight

The double-angle tangent identity is not just an exam line — it shows up wherever a rate, a slope, or an angle gets doubled or compared.

• Calculus integration. The tangent half-angle (Weierstrass) substitution $t = \tan\frac{x}{2}$ is the inverse move of this formula, and it converts trig integrals into rational ones. Knowing $\tan 2x$ in terms of $\tan x$ is the conceptual key to that substitution.

• Optics and reflection. When a mirror rotates by an angle $x$, the reflected ray rotates by $2x$ — the doubling that drives galvanometer mirrors and laser-scanning systems. Tracking the reflected slope means tracking $\tan 2x$.

• Surveying and slope geometry. Computing the angle a doubled gradient makes with the horizontal uses the same identity — the slope of a line at angle $x$ is $\tan x$, and the doubled-angle direction is $\tan 2x$.

• Signal processing. Frequency-doubling in nonlinear systems produces second-harmonic terms whose phase relationships are governed by double-angle identities, tangent included.

For Class 11 and high-school students, the everyday use is exact-value computation and proving identities — but the same formula reappears in JEE-level integration and in the rotation math behind optical instruments.

Tripping Points to Avoid

Mistake 1: Dropping the denominator

Where it slips in: Writing $\tan 2x = 2\tan x$, treating tangent as if it distributed over doubling the way $2x$ distributes.

Don't do this: Assume doubling the angle doubles the tangent. It doesn't — the $1 - \tan^2 x$ denominator is the whole point.

The correct way: Always carry the full fraction $\frac{2\tan x}{1 - \tan^2 x}$.

Mistake 2: Confusing tan2x with tan²x

Where it slips in: Reading $\tan^2 x$ (tangent squared) as $\tan 2x$ (tangent of double the angle), or the reverse.

Don't do this: Treat the two as interchangeable. $\tan^2 x = (\tan x)^2$ is a number squared; $\tan 2x$ is a different angle's tangent.

The correct way: Read the notation precisely. $\tan 2x = \frac{2\tan x}{1 - \tan^2 x}$, while $\tan^2 x$ is just $(\tan x)^2$. The memorizer archetype — fluent at both formulas — is the one who swaps them under exam pressure, because the two strings look almost identical on the page.

Mistake 3: Forgetting the formula can be undefined

Where it slips in: Plugging in $x = 45°$ (or $\frac{\pi}{4}$) and reporting a finite number instead of "undefined."

Don't do this: Force a value through a zero denominator. $\frac{2}{0}$ is not a number.

The correct way: Check the denominator before reporting an answer. If $\tan x = \pm 1$, then $\tan 2x$ is undefined — the formula is telling you about a vertical asymptote, not failing.

Conclusion

• The tan2x formula is $\tan 2x = \frac{2\tan x}{1 - \tan^2 x}$ — the double-angle identity for tangent.

• It derives in one step from the tangent sum formula with $A = B = x$, or from dividing $\sin 2x$ by $\cos 2x$.

• The denominator $1 - \tan^2 x$ is where the formula goes undefined: at $\tan x = \pm 1$ ($x = 45°$, $\frac{\pi}{4}$).

• The two most common errors are dropping the denominator ($\tan 2x \neq 2\tan x$) and confusing $\tan 2x$ with $\tan^2 x$.

• The identity underlies the Weierstrass integration substitution, mirror-rotation optics, and second-harmonic signal analysis.

Practice These Before Moving On

1. Given $\tan x = \frac{1}{2}$, find $\tan 2x$.

2. Given $\sin x = \frac{5}{13}$ with $x$ acute, find $\tan 2x$.

3. Show that $\tan 2x$ is undefined at $x = 135°$ ($\frac{3\pi}{4}$).

If Problem 2 used $\tan x = \frac{5}{13}$, return to Mistake — sine is not tangent; find $\cos x$ first.

Want a live Bhanzu trainer to walk your child through double-angle identities and the Class 11 trigonometry chapter? Book a free demo class — online globally.