What Is The Tangent Formula?
The tangent formula is the rule that defines $\tan(\theta)$ for any angle $\theta$ — at its core, the ratio of the opposite side to the adjacent side in a right triangle, or equivalently $\sin(\theta) / \cos(\theta)$ on the unit circle. The same idea then branches into a family of identities — sum, difference, double, half, and reciprocal — that let you compute or simplify tangent for almost any angle you meet.
If you only need the headline ratio, that is it. If you are mid-problem and need a specific identity, the tables below have every tangent formula in one place.
The Tangent Formula - Full Identity List
1. Basic definition
$$\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}$$
$$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$$
2. Variable key
Symbol | Meaning |
|---|---|
$\theta$ | The angle being measured (in degrees or radians) |
Opposite | Side of the right triangle directly across from $\theta$ |
Adjacent | Side of the right triangle next to $\theta$ (not the hypotenuse) |
$\sin(\theta)$ | $y$-coordinate of the point on the unit circle at angle $\theta$ |
$\cos(\theta)$ | $x$-coordinate of the same point |
$A, B$ | Two arbitrary angles in sum/difference identities |
3. Reciprocal and quotient identities
$$\tan(\theta) = \frac{1}{\cot(\theta)} \qquad \cot(\theta) = \frac{1}{\tan(\theta)}$$
$$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}$$
4. Pythagorean identity
$$1 + \tan^2(\theta) = \sec^2(\theta)$$
This drops out of $\sin^2 + \cos^2 = 1$ if you divide every term by $\cos^2(\theta)$.
5. Sum and difference
$$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$
$$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$
6. Double angle
$$\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}$$
7. Half angle
$$\tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos\theta}{\sin\theta} = \frac{\sin\theta}{1 + \cos\theta} = \pm\sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}}$$
8. Negative angle
$$\tan(-\theta) = -\tan(\theta)$$
Tangent is an odd function, which is what makes the graph symmetric about the origin.
9. Periodicity
$$\tan(\theta + \pi) = \tan(\theta)$$
The period of tangent is $\pi$ (or 180°), not $2\pi$. That single fact catches students out more than any other.
Tangent Values For Special Angles
Both degrees and radians, side by side. Worth memorising before any trig exam.
Angle (°) | Angle (rad) | $\tan(\theta)$ |
|---|---|---|
0° | $0$ | $0$ |
30° | $\pi/6$ | $\dfrac{1}{\sqrt{3}}$ (≈ 0.577) |
45° | $\pi/4$ | $1$ |
60° | $\pi/3$ | $\sqrt{3}$ (≈ 1.732) |
90° | $\pi/2$ | undefined |
120° | $2\pi/3$ | $-\sqrt{3}$ |
135° | $3\pi/4$ | $-1$ |
150° | $5\pi/6$ | $-\dfrac{1}{\sqrt{3}}$ |
180° | $\pi$ | $0$ |
270° | $3\pi/2$ | undefined |
360° | $2\pi$ | $0$ |
Where Tan Comes From - The Intuitive Derivation
A formula presented without a derivation reads as a fact to memorise. Tangent is one of the easier ones to actually see, and seeing it makes every identity downstream cheaper.
Start with the right triangle. Pick any angle $\theta$ that is not $90°$. The side opposite $\theta$ has length $y$. The side adjacent to $\theta$ has length $x$. The ratio $y/x$ is what we call $\tan(\theta)$. Nothing more.
Now place the same angle inside a unit circle, with one side along the positive $x$-axis. The point where the angle's other side hits the circle has coordinates $(\cos\theta, \sin\theta)$. Drop a vertical line from that point to the $x$-axis and you have a right triangle whose vertical leg is $\sin\theta$ and horizontal leg is $\cos\theta$. The same ratio — opposite over adjacent — is now $\sin\theta / \cos\theta$.
That is why both definitions agree, and why tan is undefined at $90°$: the horizontal leg ($\cos 90°$) is zero, and you cannot divide by zero. The graph of $\tan\theta$ shoots off to infinity at exactly those points, and that is not a quirk of the function — it is geometry refusing to let you divide by nothing.
Worked Examples of Tangent Formulas
Example 1: Find tan(75°) using the sum formula
Problem: Compute $\tan(75°)$ exactly, using $75° = 45° + 30°$.
Apply the sum formula:
$$\tan(45° + 30°) = \frac{\tan 45° + \tan 30°}{1 - \tan 45° \cdot \tan 30°}$$
Substitute known values: $\tan 45° = 1$, $\tan 30° = 1/\sqrt{3}$.
$$= \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$$
Rationalise by multiplying numerator and denominator by $(\sqrt{3} + 1)$:
$$= \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3 + 2\sqrt{3} + 1}{3 - 1} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$$
Final answer: $\tan(75°) = 2 + \sqrt{3}$.
Example 2: The wrong path, then the right one
This one shows a trap that the rusher walks into.
Problem: Compute $\tan(2 \cdot 30°)$, that is, $\tan(60°)$, using the double-angle formula.
The intuitive-but-wrong move is to write $\tan(2\theta) = 2\tan(\theta)$, by analogy with $2 \cdot 30 = 60$. So the rusher writes:
$$\tan(60°) \stackrel{?}{=} 2 \cdot \tan(30°) = 2 \cdot \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \approx 1.155$$
Check this against the table above: $\tan(60°) = \sqrt{3} \approx 1.732$. Not even close. So $\tan(2\theta) \neq 2\tan\theta$ — the function does not scale linearly with the angle.
The correct move uses the actual double-angle identity:
$$\tan(2 \cdot 30°) = \frac{2\tan(30°)}{1 - \tan^2(30°)}$$
Substitute $\tan(30°) = 1/\sqrt{3}$:
$$= \frac{2 \cdot \frac{1}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} = \frac{2}{\sqrt{3}} \cdot \frac{3}{2} = \frac{3}{\sqrt{3}} = \sqrt{3}$$
Final answer: $\tan(60°) = \sqrt{3}$.
Example 3: Real measurement — angle of elevation
Problem: From 50 metres away, the angle of elevation to the top of a tower is $32°$. How tall is the tower?
Set up the right triangle. The 50 m horizontal distance is adjacent to the $32°$ angle. The tower height $h$ is opposite.
$$\tan(32°) = \frac{h}{50}$$
Solve for $h$:
$$h = 50 \cdot \tan(32°) = 50 \cdot 0.6249 \approx 31.2$$
Final answer: The tower is approximately $31.2$ metres tall.
This is exactly the calculation surveyors use when they cannot stand at the base of what they are measuring.
Why Tangent Exists - The WHY Behind The Formula
Sine and cosine answer where a point sits on a circle. Tangent answers a different question: how steep is this thing? That is why the same formula governs ramps, roof pitches, the angle of a satellite dish, the slope of a road, and the trajectory of a projectile.
Three reasons tangent earned its own name instead of just being written as $\sin/\cos$ every time:
Slope. A slope of 1 means a $45°$ line. A slope of $\sqrt{3}$ means $60°$. The tangent of an angle is the slope of the line drawn at that angle — no conversion needed.
Surveying and astronomy. Greek and Indian astronomers needed a quick way to convert "angle to a star" into "horizontal distance per unit of vertical rise." That ratio is tangent. The names tangens (Latin for "touching") and the Sanskrit jya trace back to exactly this need.
Calculus, eventually. The derivative of any smooth function at a point is the tangent of the angle the curve makes with the horizontal. Every "instantaneous rate of change" your child meets in Grade 11 or 12 is a tangent in disguise.
Most students meet tangent as the third item in a list of three trigonometric ratios. It is worth knowing it was invented for a job — and is still doing that job every time GPS works, every time a phone screen rotates, every time a surveyor measures a building they cannot enter.
Mathematicians & History Behind The Concept
The story of tangent is a story of distance — measuring things you could not walk up to.
The astronomer who refused to wait for the star to come down
Mathematician: Hipparchus of Nicaea
Date and place: Around 150 BCE, Rhodes, ancient Greece.
The story: Hipparchus wanted to know how far the Moon was from the Earth. He could not, obviously, walk there. So he built the first known table of chord lengths in a circle — one for each angle — and used it to turn angles he could measure (from two cities at the same time) into distances he could not. That table is the direct ancestor of every sine, cosine, and tangent table that came after. He made the bet that geometry could replace travel, and won.
Why it matters: Every time a phone uses GPS, a satellite finds its altitude, or an architect sights a building from across a road, the same bet is being placed. The tools are different. The trick is identical.
Other names worth recognising as you keep doing trigonometry:
Aryabhata (5th century, India) — introduced the sine function as we now use it (he called it jya) in his astronomical work Aryabhatiya. The word sine in English comes from a Latin mistranslation of the Sanskrit term.
Nasir al-Din al-Tusi (13th century, Persia) — wrote the first treatise treating trigonometry as its own subject, separate from astronomy. He compiled the earliest comprehensive tables of all six trigonometric functions, including tangent.
Common Mistakes With Tangent Formulas
Treating $\tan(2\theta)$ as $2\tan(\theta)$.
Where it slips in: Double-angle problems, especially under exam time pressure.
Don't do this: $\tan(60°) = 2\tan(30°) = 2/\sqrt{3}$.
The correct way: Use $\tan(2\theta) = 2\tan\theta / (1 - \tan^2\theta)$. Tangent is not linear in the angle — doubling the angle is not the same as doubling the value.
Forgetting that tan is undefined at $90°$ and $270°$.
Where it slips in: Anywhere a problem asks for $\tan$ at one of those angles, or where simplification leads to $\cos\theta = 0$ in the denominator.
Don't do this: Write $\tan(90°) = \infty$ as a clean answer or carry it into a substitution.
The correct way: State that $\tan$ is undefined at those angles. If your working leads to a $\tan(90°)$ term, the working has gone wrong somewhere — back up and recheck.
Wrong sign of tan in the second and fourth quadrants.
Where it slips in: Problems with angles between $90°$ and $360°$, especially $\tan(120°)$, $\tan(135°)$, $\tan(330°)$.
Don't do this: Take the magnitude from the reference angle and forget the sign — writing $\tan(120°) = \sqrt{3}$ instead of $-\sqrt{3}$.
The correct way: In the second quadrant, $\sin > 0$ and $\cos < 0$, so $\tan < 0$. In the fourth quadrant, $\sin < 0$ and $\cos > 0$, so again $\tan < 0$. Tan is positive only in the first and third quadrants — the ASTC mnemonic ("All Students Take Calculus") tracks this.
Mixing up the sum formula for tan with the sum formula for sin or cos.
Where it slips in: Identity proofs, especially in CBSE Class 11 and US precalculus.
Don't do this: Write $\tan(A + B) = \tan A + \tan B$, by false analogy.
The correct way: The denominator matters. $\tan(A + B) = (\tan A + \tan B) / (1 - \tan A \tan B)$. The $1 - \tan A \tan B$ in the denominator is what keeps the identity consistent with the sin and cos versions.
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