# 3.3.5: Tangent Sum and Difference Formulas

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tangent of a sum or difference related to a set of tangent functions.

Suppose you were given two angles and asked to find the tangent of the difference of them. For example, can you compute:

$$\tan (120^{\circ} −40^{\circ} )$$

Would you just subtract the angles and then take the tangent of the result? Or is something more complicated required to solve this problem? Keep reading, and by the end of this lesson, you'll be able to calculate trig functions like the one above.

### tangent Sum and Difference Formulas

In this lesson, we want to find a formula that will make computing the tangent of a sum of arguments or a difference of arguments easier. As first, it may seem that you should just add (or subtract) the arguments and take the tangent of the result. However, it's not quite that easy.

To find the sum formula for tangent:

\begin{aligned} \tan (a+b) &=\dfrac{\sin (a+b)}{\cos (a+b)} && \text { Using } \tan \theta=\dfrac{\sin \theta}{\cos \theta} \\ &=\dfrac{\sin a \cos b+\sin b \cos a}{\cos a \cos b-\sin a \sin b} && \text { Substituting the sum formulas for sine and cosine} \\ &=\dfrac{\dfrac{\sin a \cos b+\sin b \cos a}{\cos a \cos b}}{\dfrac{\cos a \cos b-\sin a \sin b}{\cos a \cos b}} &&\text { Divide both the numerator and the denominator } \cos a \cos b\\ &=\dfrac{\dfrac{\sin a \cos b}{\cos a \cos b}+\dfrac{\sin b \cos a}{\cos a \cos b}}{\dfrac{\cos a \cos b}{\cos a \cos b}-\dfrac{\sin a \sin b}{\cos a \cos b}} && \text { Reduce each of the fractions } \\ &= \dfrac{\dfrac{\sin a}{\cos a}+\dfrac{\sin b}{\cos b}}{1-\dfrac{\sin a \sin b}{\cos a \cos b}} && \text { Substitute } \dfrac{\sin \theta}{\cos \theta}=\tan \theta\\ \tan (a+b)&= \dfrac{\tan a+\tan b}{1-\tan a \tan b} && \text { Sum formula for tangent } \end{aligned}

In conclusion, $$\tan (a+b)=\dfrac{\tan a+\tan b}{1−\tan a\tan b}$$. Substituting $$−b$$ for $$b$$ in the above results in the difference formula for tangent:

$$\tan (a−b)=\dfrac{\tan a−\tan b}{1+\tan a\tan b}$$

#### Using the tangent Difference Formula

1. Find the exact value of $$\tan 285^{\circ}$$.

Use the difference formula for tangent, with $$285^{\circ} =330^{\circ} −45^{\circ}$$

\begin{aligned} \tan \left(330^{\circ}-45^{\circ}\right) &=\dfrac{\tan 330^{\circ}-\tan 45^{\circ}}{1+\tan 330^{\circ} \tan 45^{\circ}} \\ &=\dfrac{-\dfrac{\sqrt{3}}{3}-1}{1-\dfrac{\sqrt{3}}{3} \cdot 1}=\dfrac{-3-\sqrt{3}}{3-\sqrt{3}} \\ &=\dfrac{-3-\sqrt{3}}{3-\sqrt{3}} \cdot \dfrac{3+\sqrt{3}}{3+\sqrt{3}} \\ &=\dfrac{-9-6 \sqrt{3}-3}{9-3} \\ &=\dfrac{-12-6 \sqrt{3}}{6} \\ &=-2-\sqrt{3} \end{aligned}

To verify this on the calculator, $$\tan 285^{\circ} =−3.732$$ and $$−2−\sqrt{3}=−3.732$$.

2. Verify the tangent difference formula by finding $$\tan \dfrac{6\pi}{6}$$, since this should be equal to $$\tan \pi =0$$.

Use the difference formula for tangent, with $$\tan \dfrac{6\pi}{6}=\tan \left(\dfrac{7\pi}{6}−\dfrac{\pi}{6}\right)$$

\begin{aligned} \tan \left(\dfrac{7 \pi}{6}-\dfrac{\pi}{6}\right) &=\dfrac{\tan \dfrac{7 \pi}{6}-\tan \dfrac{\pi}{6}}{1+\tan \dfrac{7 \pi}{6} \tan \dfrac{\pi}{6}} \\ &=\dfrac{\dfrac{\sqrt{2}}{6}-\dfrac{\sqrt{2}}{6}}{1-\dfrac{\sqrt{2}}{6} \cdot \dfrac{\sqrt{2}}{6}}=\dfrac{0}{1-\dfrac{2}{36}}\\ &=\dfrac{0}{\dfrac{34}{36}} \\ &=0 \end{aligned}

3. Find the exact value of $$\tan 165^{\circ}$$.

Use the difference formula for tangent, with $$165^{\circ} =225^{\circ} −60^{\circ}$$

\begin{aligned} \tan (225^{\circ} −60^{\circ} )&= \dfrac{\tan 225^{\circ} −\tan 60^{\circ}}{1+\tan 225^{\circ} \tan 60^{\circ}} \\&=\dfrac{1−\sqrt{3}}{1−1\cdot \sqrt{3}}=1 \end{aligned}

##### Example $$\PageIndex{1}$$

Earlier, you were asked to find $$\tan (120^{\circ} −40^{\circ} )$$.

Solution

You can use the tangent difference formula:

$$\tan (a−b)=\dfrac{\tan a−\tan b}{1+\tan a \tan b}$$

to help solve this. Substituting in known quantities:

$$\tan (120^{\circ} −40^{\circ} )=\dfrac{\tan 120^{\circ} −\tan 40^{\circ}}{1+(\tan 120^{\circ} )(\tan 40^{\circ} )}=\dfrac{−1.732−.839}{1+(−1.732)(.839)}=\dfrac{−2.571}{−.453148}=5.674$$

##### Example $$\PageIndex{2}$$

Find the exact value for $$\tan 75^{\circ}$$

Solution

\begin{aligned} \tan 75^{\circ} &=\tan \left(45^{\circ}+30^{\circ}\right) \\ &=\dfrac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ} \tan 30^{\circ}} \\ &=\dfrac{1+\dfrac{\sqrt{3}}{3}}{1-1 \cdot \dfrac{\sqrt{3}}{3}}=\dfrac{\dfrac{3+\sqrt{3}}{3}}{\dfrac{3-\sqrt{3}}{3}} \\ &=\dfrac{3+\sqrt{3}}{3-\sqrt{3}} \cdot \dfrac{3+\sqrt{3}}{3+\sqrt{3}} \\ &=\dfrac{9+6 \sqrt{3}+3}{9-3}=\dfrac{12+6 \sqrt{3}}{6} \\ &=2+\sqrt{3} \end{aligned}

##### Example $$\PageIndex{3}$$

Simplify $$\tan (\pi +\theta )$$

Solution

$$\tan (\pi +\theta )=\dfrac{\tan \pi +\tan \theta }{1−\tan \pi \tan \theta }=\dfrac{\tan \theta }{1}=\tan \theta$$

##### Example $$\PageIndex{4}$$

Find the exact value for $$\tan 15^{\circ}$$

Solution

\begin{aligned} \tan 15^{\circ} &=\tan \left(45^{\circ}-30^{\circ}\right) \\ &=\dfrac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}} \\ &=\dfrac{1-\dfrac{\sqrt{3}}{3}}{1+1 \cdot \dfrac{\sqrt{3}}{3}}=\dfrac{\dfrac{3-\sqrt{3}}{3}}{\dfrac{3+\sqrt{3}}{3}} \\ &=\dfrac{3-\sqrt{3}}{3+\sqrt{3}} \cdot \dfrac{3-\sqrt{3}}{3-\sqrt{3}} \\ &=\dfrac{9+6 \sqrt{3}+3}{9-3}=\dfrac{12+6 \sqrt{3}}{6} \\ &=2+\sqrt{3} \end{aligned}

## Review

Find the exact value for each tangent expression.

1. $$\tan \dfrac{5 \pi}{12}$$
2. $$\tan \dfrac{11 \pi}{12}$$
3. $$\tan −165^{\circ}$$
4. $$\tan 255^{\circ}$$
5. $$\tan −15^{\circ}$$

Write each expression as the tangent of an angle.

1. $$\dfrac{\tan 15^{\circ} +\tan 42^{\circ}}{1−\tan 15^{\circ} \tan 42^{\circ}}$$
2. $$\dfrac{\tan 65^{\circ} −\tan 12^{\circ} }{1+\tan 65^{\circ} \tan 12^{\circ}}$$
3. $$\dfrac{\tan 10^{\circ} +\tan 50^{\circ}}{1−\tan 10^{\circ} \tan 50^{\circ}}$$
4. $$\dfrac{\tan 2y+\tan 4}{1−\tan 2 \tan 4y}$$
5. $$\dfrac{\tan x−\tan 3x}{1+\tan x\tan 3x}$$
6. $$\dfrac{\tan 2x−\tan y}{1+\tan 2x\tan y}$$
7. Prove that $$\tan \left(x+\dfrac{\pi }{4}\right)=\dfrac{1+\tan (x)}{1−\tan (x)}$$
8. Prove that $$\tan \left(x−\dfrac{\pi }{2}\right)=−\cot (x)$$
9. Prove that $$\tan \left(\dfrac{\pi }{2}−x\right)=\cot (x)$$
10. Prove that $$\tan (x+y)\tan (x−y)=\dfrac{\tan^2(x)−\tan^2(y)}{1−\tan^2(x)\tan^2(y)}$$