Tan 75 Degrees: Exact Value, Proofs And Worked Examples

What Does Tan 75 Degrees Mean?

In a right triangle, tan 75 degrees is the length of the side opposite the $75^\circ$ angle divided by the side adjacent to it. On the unit circle, a circle of radius $1$ centred at the origin, it is the $y$-coordinate divided by the $x$-coordinate of the point at $75^\circ$, as the tangent function is defined.

Terms used below, defined first:

• Quadrant: one of the four regions cut by the axes, numbered Q1 to Q4 counterclockwise from the top right. $75^\circ$ is in Q1, where sine, cosine and tangent are all positive.

• Reference angle: the acute angle between the terminal side and the $x$-axis; for any Q1 angle it is the angle itself.

• Angle-sum identity for tangent: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$, one of the sum and difference identities.

Because $75^\circ$ is not one of the special angles $30^\circ$, $45^\circ$ or $60^\circ$, its exact value has to be built from them. Proving and using the addition formulas is CCSS-M standard F-TF.C.9, and NCERT Class 11 Mathematics, Chapter 3 (Trigonometric Functions), asks for the partner value $\tan 15^\circ$ in its sum-and-difference exercise.

Standard-Angle Tangent Reference Table

Tangent has exact values at every multiple of $15^\circ$, and $75^\circ$ is the last stop before tangent becomes undefined at $90^\circ$. The table pairs degrees with radians.

Two patterns stand out. The values for $15^\circ$ and $75^\circ$ are reciprocals ($2 - \sqrt{3}$ and $2 + \sqrt{3}$ multiply to $1$), and the jump from $60^\circ$ to $75^\circ$ is bigger than the whole climb from $0^\circ$ to $60^\circ$.

Where Does Tan 75 Degrees Show Up, And Why Does It Matter?

Tan 75 degrees shows up wherever a steep angle has to be turned into a ratio of heights to distances. Because tangent is rise over run, it converts a $75^\circ$ angle straight into "how far up for each step across".

• Slope: a line at $75^\circ$ to the horizontal has slope $2 + \sqrt{3} \approx 3.73$, so it rises about $3.73$ units for every unit across.

• Heights: when the angle of elevation to a treetop is $75^\circ$ from $d$ metres away, the tree is $d \tan 75^\circ$ metres above eye level.

• Regular 12-sided shapes: a regular dodecagon with side $1$ has apothem (centre-to-side distance) $\frac{1}{2}\tan 75^\circ = \frac{2 + \sqrt{3}}{2} \approx 1.8660$, because each side subtends $30^\circ$ at the centre.

What Is Tan 75 Degrees In Fraction Form?

As a fraction, tan 75 degrees comes out first as $\frac{\sqrt{3} + 1}{\sqrt{3} - 1}$, and rationalising turns it into $2 + \sqrt{3}$. That rationalised form is the "root form" textbooks expect.

$$\frac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} + 1)^2}{3 - 1}$$

$$= \frac{3 + 2\sqrt{3} + 1}{2} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$$

No ordinary fraction of whole numbers equals $2 + \sqrt{3}$, because $\sqrt{3}$ is irrational. Fractions such as $\frac{97}{26} \approx 3.7308$ only approximate it.

How Do You Find Tan 75 Degrees?

Write $75^\circ$ as a sum or difference of angles with known tangents, then apply the matching identity. Each route below gives $2 + \sqrt{3}$.

Method 1: The Sum Formula, $\tan(45^\circ + 30^\circ)$

Use tan 45 degrees $= 1$ and $\tan 30^\circ = \frac{1}{\sqrt{3}}$:

$$\tan 75^\circ = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ}$$

$$= \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}}$$

Multiply the top and bottom by $\sqrt{3}$:

$$= \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = 2 + \sqrt{3}$$

The last step is the rationalisation shown in the fraction-form section.

Method 2: The Difference Formula, $\tan(120^\circ - 45^\circ)$

The tan(A − B) formula works with $A = 120^\circ$, where tan 120 degrees $= -\sqrt{3}$:

$$\tan 75^\circ = \frac{\tan 120^\circ - \tan 45^\circ}{1 + \tan 120^\circ \tan 45^\circ}$$

$$= \frac{-\sqrt{3} - 1}{1 - \sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} = 2 + \sqrt{3}$$

Method 3: The Half-Angle Formula, $\tan\left(\frac{150^\circ}{2}\right)$

$75^\circ$ is half of $150^\circ$, with $\sin 150^\circ = \frac{1}{2}$ and $\cos 150^\circ = -\frac{\sqrt{3}}{2}$. The half angle formula for tangent, $\tan \frac{\theta}{2} = \frac{1 - \cos\theta}{\sin\theta}$, gives

$$\tan 75^\circ = \frac{1 - \left(-\frac{\sqrt{3}}{2}\right)}{\frac{1}{2}} = \frac{1 + \frac{\sqrt{3}}{2}}{\frac{1}{2}}$$

$$= 2 + \sqrt{3}$$

This form has no $\pm$ sign and needs no rationalising, which makes it the quickest route once the formula is known.

Method 4: Sine Over Cosine

Divide the exact values of sin 75 degrees and $\cos 75^\circ$:

$$\tan 75^\circ = \frac{\sin 75^\circ}{\cos 75^\circ} = \frac{\frac{\sqrt{6} + \sqrt{2}}{4}}{\frac{\sqrt{6} - \sqrt{2}}{4}} = \frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}}$$

$$= \frac{(\sqrt{6} + \sqrt{2})^2}{6 - 2} = \frac{8 + 4\sqrt{3}}{4} = 2 + \sqrt{3}$$

Method 5: The Co-Function Route, $\cot 15^\circ$

Complementary angles swap tangent and cotangent, as the trigonometric ratios of complementary angles show: $\tan 75^\circ = \cot 15^\circ$. With $\tan 15^\circ = 2 - \sqrt{3}$,

$$\tan 75^\circ = \frac{1}{2 - \sqrt{3}} \times \frac{2 + \sqrt{3}}{2 + \sqrt{3}} = \frac{2 + \sqrt{3}}{4 - 3} = 2 + \sqrt{3}$$

Method 6: The 15-75-90 Triangle (No Formula Needed)

Start from a 30-60-90 triangle ABC with the right angle at C, $\angle A = 30^\circ$, $BC = 1$, $AC = \sqrt{3}$ and $AB = 2$. Extend side CA past A to a point D with $AD = AB = 2$.

Triangle ABD is isosceles, so $\angle ADB = \angle ABD$, and together they equal the exterior angle $\angle BAC = 30^\circ$. That makes $\angle ADB = 15^\circ$ and, in the large right triangle DCB, $\angle DBC = 90^\circ - 15^\circ = 75^\circ$.

$$\tan \angle DBC = \frac{DC}{BC} = \frac{2 + \sqrt{3}}{1} = 2 + \sqrt{3}$$

Method 7: Calculator Check

In degree mode, $\tan(75)$ returns $3.7320508\ldots$, and $2 + \sqrt{3} = 2 + 1.7320508\ldots = 3.7320508\ldots$, which matches.

Why Is Tan 75 Degrees Bigger Than 3?

Tan 75 degrees is bigger than $3$ because tangent grows faster and faster as the angle approaches $90^\circ$. At $75^\circ$ the adjacent side has shrunk to about a quarter of the opposite side, so their ratio is close to $4$.

The table shows it: tangent goes $1$ at $45^\circ$, $1.73$ at $60^\circ$, then $3.73$ at $75^\circ$, and it has no value at all at $90^\circ$. So any answer for tan 75 degrees that is smaller than $\tan 60^\circ \approx 1.73$ must be wrong, a check worth running after every calculation.

Why Does A Calculator Show -0.4207 For Tan 75?

A calculator in radian mode reads "tan 75" as the tangent of $75$ radians, about $11.9$ full turns, and returns $-0.4207$. Google's search bar does the same for the unbracketed query "tan 75", because it treats a bare number as radians.

Typing "tan 75 degrees" or switching to degree mode gives $3.7321$. The reverse slip also happens: entering $\frac{5\pi}{12}$ in degree mode returns $\tan(1.309^\circ) \approx 0.0229$.

Examples Of Tan 75 Degrees

Example 1

Evaluate $2\tan 75^\circ$.

$$2\tan 75^\circ = 2(2 + \sqrt{3})$$

$$= 4 + 2\sqrt{3} \approx 7.4641$$

Example 2

Find $\tan 75^\circ$ using $75^\circ = 45^\circ + 30^\circ$.

A first attempt adds the tangents:

$$\tan 45^\circ + \tan 30^\circ = 1 + \frac{\sqrt{3}}{3} \approx 1.5774$$

That fails the size check, because $\tan 75^\circ$ must be larger than $\tan 60^\circ \approx 1.7321$. Tangent does not split across a sum; the identity divides by $1 - \tan A \tan B$:

$$\tan 75^\circ = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$$

Final answer: $2 + \sqrt{3} \approx 3.7321$.

Example 3

Show that $\tan 75^\circ + \cot 75^\circ = 4$.

$\cot 75^\circ = \frac{1}{\tan 75^\circ} = \frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3}$ after rationalising.

$$\tan 75^\circ + \cot 75^\circ = (2 + \sqrt{3}) + (2 - \sqrt{3})$$

$$= 4$$

Example 4

Find the value of $\tan 75^\circ - \cot 75^\circ$.

$$\tan 75^\circ - \cot 75^\circ = (2 + \sqrt{3}) - (2 - \sqrt{3})$$

$$= 2\sqrt{3} \approx 3.4641$$

Example 5

Find $\tan^2 75^\circ$ in exact form, and use it to find $\sec^2 75^\circ$.

$$\tan^2 75^\circ = (2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3$$

$$= 7 + 4\sqrt{3} \approx 13.9282$$

By the identity $\sec^2\theta = 1 + \tan^2\theta$, $\sec^2 75^\circ = 8 + 4\sqrt{3} \approx 14.9282$. Students who square $2 + \sqrt{3}$ as $4 + 3 = 7$ usually drop the middle term $2 \times 2 \times \sqrt{3}$; writing the square as two brackets multiplied out makes it visible.

Example 6

From a point $20$ m from the base of a tower, the angle of elevation to the top is $75^\circ$. How tall is the tower above eye level?

$$h = 20 \tan 75^\circ = 20(2 + \sqrt{3})$$

$$= 40 + 20\sqrt{3} \approx 74.64 \text{ m}$$

Which Mistakes Cost Marks On Tan 75 Degrees?

Mistake 1: Adding The Tangents Of $45^\circ$ And $30^\circ$

Where it slips in: The first line after writing $75^\circ = 45^\circ + 30^\circ$.

Don't do this: Write $\tan 75^\circ = 1 + \frac{1}{\sqrt{3}} \approx 1.5774$.

The correct way: Use $\frac{\tan A + \tan B}{1 - \tan A \tan B}$. The rusher who distributes "tan" like a multiplier gets a value below $\tan 60^\circ$; comparing every answer with the neighbouring table values catches it at once.

Mistake 2: Using A Plus Sign In The Denominator

Where it slips in: Recalling the sum formula from memory, mixing it up with the difference formula.

Don't do this: Write $\frac{1 + \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = 1$, which would make $75^\circ$ behave like $45^\circ$.

The correct way: The sum formula has a minus below and a plus above; the difference formula flips both. The memoriser who stores the two formulas as one picture tends to blur these signs, and the habit that fixes it is deriving the tangent formula once from $\frac{\sin(A + B)}{\cos(A + B)}$.

Mistake 3: Leaving The Calculator In Radian Mode

Where it slips in: Checking the exact answer on a phone or a search bar.

Don't do this: Accept $-0.4207$, which is the tangent of $75$ radians.

The correct way: Set degree mode or type "tan 75 degrees". A negative tangent for a Q1 angle is impossible, so a minus sign on the screen means the mode is wrong.

Where Can You Get Help With Trigonometry Formulas?

Tan 75 degrees is often the first exact value students build with a formula rather than read from a triangle. A trigonometry tutor or a high school math tutor can practise the sum, difference and half-angle routes with you, and regular math tutoring keeps those identities fresh.

Key Takeaways

• Tan 75 degrees equals $2 + \sqrt{3}$, about $3.7321$; in radians it is $\tan \frac{5\pi}{12}$.

• The standard proof is $\tan(45^\circ + 30^\circ) = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}}$, then rationalising.

• The half-angle route $\frac{1 - \cos 150^\circ}{\sin 150^\circ}$ reaches $2 + \sqrt{3}$ with no rationalising.

• Tan 75 degrees equals $\cot 15^\circ$, and $\tan 75^\circ \times \tan 15^\circ = 1$.

• Any answer below $\tan 60^\circ \approx 1.73$, or any negative answer, signals an error.

• A calculator must be in degree mode; radian mode returns $-0.4207$.

Practice These Before Moving On

1. Find $\tan 15^\circ$ exactly using $\tan(45^\circ - 30^\circ)$.

2. Find $\tan 105^\circ$.

3. Evaluate $\tan 75^\circ \times \tan 15^\circ$.

4. Find $\tan 165^\circ$.

Answers:

1. $\frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = 2 - \sqrt{3} \approx 0.2679$.

2. $105^\circ = 180^\circ - 75^\circ$, and tangent is negative in Q2, so $\tan 105^\circ = -(2 + \sqrt{3}) \approx -3.7321$.

3. $(2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$.

4. $165^\circ = 180^\circ - 15^\circ$, so $\tan 165^\circ = -(2 - \sqrt{3}) = \sqrt{3} - 2 \approx -0.2679$.

If question 1 came out as $2 + \sqrt{3}$, check the signs in Mistake 2: the difference formula flips both of them.

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Read More

• Tan 60 degrees and the 30-60-90 triangle

• Tan 30 degrees in exact form

• Cos 15 degrees and its exact value

• Heights and distances with trigonometry

• The trigonometric table of standard angles