# Addition and Subtraction Formulas for Tangent and Cotangent

### Tangent Addition Formula

On the previous page, we have derived the addition identities for sine and cosine:

${\sin \left( {\alpha + \beta } \right) }={ \sin \alpha \cos \beta + \cos \alpha \sin \beta ,}$

${\cos \left( {\alpha + \beta } \right) }={ \cos \alpha \cos \beta – \sin \alpha \sin \beta .}$

Suppose now that $$\cos \left( {\alpha + \beta } \right) \ne 0,$$ or $$\alpha + \beta \ne \frac{\pi }{2} + \pi n,$$ $$n \in \mathbb{Z}.$$ Moreover, let also $$\cos \alpha \ne 0$$ and $$\cos \beta \ne 0,$$ that is, $$\alpha, \beta \ne \frac{\pi }{2} + \pi n,$$ $$n \in \mathbb{Z},$$ so that we can divide by $$\cos\alpha\cos\beta.$$

Then the tangent addition formula is given by

$\require{cancel}{\tan \left( {\alpha + \beta } \right) }={ \frac{{\sin \left( {\alpha + \beta } \right)}}{{\cos \left( {\alpha + \beta } \right)}} }={ \frac{{\sin \alpha \cos \beta + \cos \alpha \sin \beta }}{{\cos \alpha \cos \beta – \sin \alpha \sin \beta }} }={ \frac{{\frac{{\sin \alpha \cos \beta + \cos \alpha \sin \beta }}{{\cos \alpha \cos \beta }}}}{{\frac{{\cos \alpha \cos \beta – \sin \alpha \sin \beta }}{{\cos \alpha \cos \beta }}}} }={ \frac{{\frac{{\sin \alpha \cancel{\cos \beta} }}{{\cos \alpha \cancel{\cos \beta} }} + \frac{{\cancel{\cos \alpha} \sin \beta }}{{\cancel{\cos \alpha} \cos \beta }}}}{{\frac{\cancel{\cos \alpha \cos \beta }}{\cancel{\cos \alpha \cos \beta }} – \frac{{\sin \alpha \sin \beta }}{{\cos \alpha \cos \beta }}}} }={ \frac{{\tan \alpha + \tan \beta }}{{1 – \tan \alpha \tan \beta }}.}$

Hence,

### Tangent Subtraction Formula

The tangent function is odd:

${\tan \left( { – \beta } \right) }={ \frac{{\sin \left( { – \beta } \right)}}{{\cos \left( { – \beta } \right)}} }={ \frac{{ – \sin \beta }}{{\cos \beta }} }={ – \tan \beta .}$

Replacing $$\beta \to -\beta$$ in the tangent addition formula, we obtain the tangent subtraction formula:

${\tan \left( {\alpha – \beta } \right) }={ \frac{{\tan \alpha + \tan \left( { – \beta } \right)}}{{1 – \tan \alpha \tan \left( { – \beta } \right)}} }={ \frac{{\tan \alpha – \tan \beta }}{{1 + \tan \alpha \tan \beta }}.}$

Thus,

### Cotangent Addition Formula

Similarly we can establish the addition identity for cotangent.

Let $$\sin \left( {\alpha + \beta } \right) \ne 0,$$ that is, $$\alpha + \beta \ne \pi n,$$ $$n \in \mathbb{Z}.$$ We also assume that $$\sin\alpha \ne 0$$ and $$\sin\beta \ne 0,$$ or $$\alpha ,\beta \ne \pi n,$$ $$n \in \mathbb{Z},$$ so that we can divide by $$\sin\alpha\sin\beta.$$

Then we have

${\cot \left( {\alpha + \beta } \right) }={ \frac{{\cos \left( {\alpha + \beta } \right)}}{{\sin \left( {\alpha + \beta } \right)}} }={ \frac{{\cos \alpha \cos \beta – \sin \alpha \sin \beta }}{{\sin \alpha \cos \beta + \cos \alpha \sin \beta }} }={ \frac{{\frac{{\cos \alpha \cos \beta – \sin \alpha \sin \beta }}{{\sin \alpha \sin \beta }}}}{{\frac{{\sin \alpha \cos \beta + \cos \alpha \sin \beta }}{{\sin \alpha \sin \beta }}}} }={ \frac{{\frac{{\cos \alpha \cos \beta }}{{\sin \alpha \sin \beta }} – \frac{\cancel{\sin \alpha \sin \beta }}{\cancel{\sin \alpha \sin \beta }}}}{{\frac{{\cancel{\sin \alpha} \cos \beta }}{{\cancel{\sin \alpha} \sin \beta }} + \frac{{\cos \alpha \cancel{\sin \beta} }}{{\sin \alpha \cancel{\sin \beta} }}}} }={ \frac{{\cot \alpha \cot \beta – 1}}{{\cot \beta + \cot \alpha }}.}$

We got the following result:

The cotangent of the sum of two angles can also be expressed in terms of tangents:

### Cotangent Subtraction Formula

First we note that the cotangent function is odd:

${\cot \left( { – \alpha } \right) }={ \frac{{\cos \left( { – \alpha } \right)}}{{\sin \left( { – \alpha } \right)}} }={ \frac{{\cos \alpha }}{{ – \sin \alpha }} }={ – \cot \alpha .}$

Now we can easily derive the cotangent subtraction formula. It is obtained by replacing $$\beta \to -\beta$$ in the cotangent addition formula:

${\cot \left( {\alpha – \beta } \right) }={ \frac{{\cot \alpha \cot \left( { – \beta } \right) – 1}}{{\cot \alpha + \cot \left( { – \beta } \right)}} }={ \frac{{ – \cot \alpha \cot \beta – 1}}{{\cot \alpha – \cot \beta }} }={ \frac{{\cot \alpha \cot \beta + 1}}{{\cot \beta – \cot \alpha }}.}$

So, we have

In terms of tangents, the cotangent subtraction formula is given by

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Calculate $$\tan \large{\frac{{5\pi }}{{12}}}\normalsize.$$

### Example 2

Calculate $$\cot 15^\circ.$$

### Example 3

Find the value of $$\tan \left( {\large{\frac{\pi }{3}}\normalsize + \alpha } \right)$$ if $$\cos \alpha = 0.6$$ and the angle $$\alpha$$ lies in the $$4\text{th}$$ quadrant.

### Example 4

Find the value of $$\cot \left( {\large{\frac{\pi }{4}}\normalsize – \beta } \right)$$ if $$\sin \beta = -0.8$$ and the angle $$\beta$$ lies in the $$3\text{rd}$$ quadrant.

### Example 5

Simplify the expression $\frac{{2\tan \alpha }}{{1 – {{\tan }^2}\alpha }}\cos 2\alpha – \sin 2\alpha .$

### Example 6

Simplify the expression $\frac{{\tan \frac{{31\pi }}{{36}} – \tan \frac{\pi }{9}}}{{1 – \tan \frac{{5\pi }}{{36}}\tan \frac{\pi }{9}}}.$

### Example 7

Calculate $\frac{{{{\tan }^2}\frac{{7\pi }}{{24}} – {{\tan }^2}\frac{\pi }{{24}}}}{{1 – {{\tan }^2}\frac{{7\pi }}{{24}}\,{{\tan }^2}\frac{\pi }{{24}}}}.$

### Example 8

Prove that $\tan \left( {\alpha + \beta } \right) \gt \tan \alpha + \tan \beta$ if $$0 \lt \alpha \lt \large{\frac{\pi }{4}}\normalsize$$ and $$0 \lt \beta \lt \large{\frac{\pi }{4}}\normalsize.$$

### Example 9

Find the tangent of the angle at which the curves $$y_1 = x^2$$ and $$y_2 = \large{\frac{1}{x}}\normalsize$$ intersect.

### Example 10

Find the tangent of the angle of intersection of the curves $$y_1 = \ln x + 1$$ and $$y_2 = \large{\frac{1}{x^2}}\normalsize.$$

### Example 1.

Calculate $$\tan \large{\frac{{5\pi }}{{12}}}\normalsize.$$

Solution.

We write the angle $$\large{\frac{{5\pi }}{{12}}}\normalsize$$ as the sum of two special angles:

${\frac{{5\pi }}{{12}} = \frac{{3\pi + 2\pi }}{{12}} }={ \frac{{3\pi }}{{12}} + \frac{{2\pi }}{{12}} }={ \frac{\pi }{4} + \frac{\pi }{6}.}$

Apply the tangent addition identity:

${\tan \frac{{5\pi }}{{12}} = \tan \left( {\frac{\pi }{4} + \frac{\pi }{6}} \right) }={ \frac{{\tan \frac{\pi }{4} + \tan \frac{\pi }{6}}}{{1 – \tan \frac{\pi }{4}\tan \frac{\pi }{6}}} }={ \frac{{1 + \frac{1}{{\sqrt 3 }}}}{{1 – 1 \cdot \frac{1}{{\sqrt 3 }}}} }={ \frac{{\sqrt 3 + 1}}{{\sqrt 3 – 1}} }={ \frac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{{\left( {\sqrt 3 – 1} \right)\left( {\sqrt 3 + 1} \right)}} }={ \frac{{3 + 2\sqrt 3 + 1}}{{3 – 1}} }={ \frac{{4 + 2\sqrt 3 }}{2} }={ 2 + \sqrt 3 .}$

### Example 2.

Calculate $$\cot 15^\circ.$$

Solution.

We represent the angle $$15^\circ$$ as the difference of two angles: $$15^\circ = 45^\circ – 30^\circ.$$ Then we have

${\cot {15^\circ} = \cot \left( {{{45}^\circ} – {{30}^\circ}} \right) }={ \frac{{\cot {{45}^\circ}\cot {{30}^\circ} + 1}}{{\cot {{30}^\circ} – \cot {{45}^\circ}}} }={ \frac{{1 \cdot \sqrt 3 + 1}}{{\sqrt 3 – 1}} }={ \frac{{\sqrt 3 + 1}}{{\sqrt 3 – 1}} }={ \frac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{{\left( {\sqrt 3 – 1} \right)\left( {\sqrt 3 + 1} \right)}} }={ \frac{{3 + 2\sqrt 3 + 1}}{{3 – 1}} }={ \frac{{4 + 2\sqrt 3 }}{2} }={ 2 + \sqrt 3 .}$

### Example 3.

Find the value of $$\tan \left( {\large{\frac{\pi }{3}}\normalsize + \alpha } \right)$$ if $$\cos \alpha = 0.6$$ and the angle $$\alpha$$ lies in the $$4\text{th}$$ quadrant.

Solution.

First we determine $$\sin\alpha$$ using the Pythagorean trigonometric identity. The sine has a negative value because $$\alpha$$ is in the $$4\text{th}$$ quadrant.

${\sin \alpha }={ – \sqrt {1 – {{\cos }^2}\alpha } }={ – \sqrt {1 – {{0.6}^2}} }={ – \sqrt {1 – 0.36} }={ – \sqrt {0.64} }={ – 0.8}$

Hence, the tangent is equal to

${\tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} }={ \frac{{ – 0.8}}{{0.6}} }={ – \frac{4}{3}.}$

Now we can compute $$\tan \left( {\large{\frac{\pi }{3}}\normalsize + \alpha } \right):$$

${\tan \left( {\frac{\pi }{3} + \alpha } \right) }={ \frac{{\tan \frac{\pi }{3} + \tan \alpha }}{{1 – \tan \frac{\pi }{3}\tan \alpha }} }={ \frac{{\sqrt 3 – \frac{4}{3}}}{{1 – \sqrt 3 \cdot \left( { – \frac{4}{3}} \right)}} }={ \frac{{3\sqrt 3 – 4}}{{4\sqrt 3 + 3}} }={ \frac{{\left( {3\sqrt 3 – 4} \right)\left( {4\sqrt 3 – 3} \right)}}{{\left( {4\sqrt 3 + 3} \right)\left( {4\sqrt 3 – 3} \right)}} }={ \frac{{36 – 16\sqrt 3 – 9\sqrt 3 + 12}}{{48 – 9}} }={ \frac{{48 – 25\sqrt 3 }}{{39}}.}$

### Example 4.

Find the value of $$\cot \left( {\large{\frac{\pi }{4}}\normalsize – \beta } \right)$$ if $$\sin \beta = -0.8$$ and the angle $$\beta$$ lies in the $$3\text{rd}$$ quadrant.

Solution.

Calculate the cosine and cotangent of $$\beta.$$ In quadrant $$III,$$ the cosine is negative. So, we have

${\cos \beta }={ – \sqrt {1 – {{\sin }^2}\beta } }={ – \sqrt {1 – {{\left( { – 0.8} \right)}^2}} }={ – \sqrt {1 – 0.64} }={ – \sqrt {0.36} }={ – 0.6}$

Hence,

${\cot \beta = \frac{{\cos \beta }}{{\sin \beta }} }={ \frac{{ – 0.6}}{{ – 0.8}} }={ \frac{3}{4}.}$

By the cotangent subtraction formula,

${\cot \left( {\frac{\pi }{4} – \beta } \right) }={ \frac{{\cot \frac{\pi }{4}\cot \beta + 1}}{{\cot \beta – \cot \frac{\pi }{4}}} }={ \frac{{1 \cdot \frac{3}{4} + 1}}{{\frac{3}{4} – 1}} }={ \frac{{\frac{7}{4}}}{{ – \frac{1}{4}}} }={ – 7.}$

### Example 5.

Simplify the expression $\frac{{2\tan \alpha }}{{1 – {{\tan }^2}\alpha }}\cos 2\alpha – \sin 2\alpha .$

Solution.

Note that

${\frac{{2\tan \alpha }}{{1 – {{\tan }^2}\alpha }} }={ \frac{{\tan \alpha + \tan \alpha }}{{1 – \tan \alpha \tan \alpha }} }={ \tan \left( {\alpha + \alpha } \right) }={ \tan 2\alpha .}$

Then

$\require{cancel}{\frac{{2\tan \alpha }}{{1 – {{\tan }^2}\alpha }}\cos 2\alpha – \sin 2\alpha }={ \tan 2\alpha \cos 2\alpha – \sin 2\alpha }={ \frac{{\sin 2\alpha \cancel{\cos 2\alpha} }}{\cancel{\cos 2\alpha }} – \sin 2\alpha }={ \sin 2\alpha – \sin 2\alpha }={ 0.}$

When dividing by $$\cos2\alpha$$ we assume that

${\cos 2\alpha \ne 0,}\;\; \Rightarrow {2\alpha \ne \frac{\pi }{2} + \pi n,}\;\; \Rightarrow {\alpha \ne \frac{\pi }{4} + \frac{{\pi n}}{2},\;}\kern0pt{n \in \mathbb{Z}.}$

### Example 6.

Simplify the expression $\frac{{\tan \frac{{31\pi }}{{36}} – \tan \frac{\pi }{9}}}{{1 – \tan \frac{{5\pi }}{{36}}\tan \frac{\pi }{9}}}.$

Solution.

We rewrite the first term in the numerator in the form

${\tan \frac{{31\pi }}{{36}} = \tan \frac{{36\pi – 5\pi }}{{36}} }={ \tan \left( {\pi – \frac{{5\pi }}{{36}}} \right).}$

The tangent is an odd function, with a period of $$\pi.$$ Therefore

${\tan \frac{{31\pi }}{{36}} = \tan \left( {\pi – \frac{{5\pi }}{{36}}} \right) }={ \tan \left( { – \frac{{5\pi }}{{36}}} \right) }={ – \tan \frac{{5\pi }}{{36}}.}$

Substitute this result into the initial expression and simplify using the tangent addition formula:

${\frac{{\tan \frac{{31\pi }}{{36}} – \tan \frac{\pi }{9}}}{{1 – \tan \frac{{5\pi }}{{36}}\tan \frac{\pi }{9}}} }={ \frac{{ – \tan \frac{{5\pi }}{{36}} – \tan \frac{\pi }{9}}}{{1 – \tan \frac{{5\pi }}{{36}}\tan \frac{\pi }{9}}} }={ – \frac{{\tan \frac{{5\pi }}{{36}} + \tan \frac{\pi }{9}}}{{1 – \tan \frac{{5\pi }}{{36}}\tan \frac{\pi }{9}}} }={ – \tan \left( {\frac{{5\pi }}{{36}} + \frac{\pi }{9}} \right) }={ – \tan \frac{{5\pi + 4\pi }}{{36}} }={ – \tan \frac{{9\pi }}{{36}} }={ – \tan \frac{\pi }{4} }={ – 1.}$

### Example 7.

Calculate $\frac{{{{\tan }^2}\frac{{7\pi }}{{24}} – {{\tan }^2}\frac{\pi }{{24}}}}{{1 – {{\tan }^2}\frac{{7\pi }}{{24}}\,{{\tan }^2}\frac{\pi }{{24}}}}.$

Solution.

We need here the difference of squares formula:

${a^2} – {b^2} = \left( {a – b} \right)\left( {a + b} \right).$

Applying this identity to both the numerator and denominator and using the tangent addition and subtraction formulas, we have

${\frac{{{{\tan }^2}\frac{{7\pi }}{{24}} – {{\tan }^2}\frac{\pi }{{24}}}}{{1 – {{\tan }^2}\frac{{7\pi }}{{24}}{{\tan }^2}\frac{\pi }{{24}}}} \text{ = }}\kern0pt{ \frac{{\left( {\tan \frac{{7\pi }}{{24}} – \tan \frac{\pi }{{24}}} \right)\left( {\tan \frac{{7\pi }}{{24}} + \tan \frac{\pi }{{24}}} \right)}}{{\left( {1 – \tan \frac{{7\pi }}{{24}}\tan \frac{\pi }{{24}}} \right)\left( {1 + \tan \frac{{7\pi }}{{24}}\tan \frac{\pi }{{24}}} \right)}} }={ \frac{{\tan \frac{{7\pi }}{{24}} – \tan \frac{\pi }{{24}}}}{{1 + \tan \frac{{7\pi }}{{24}}\tan \frac{\pi }{{24}}}} \cdot \frac{{\tan \frac{{7\pi }}{{24}} + \tan \frac{\pi }{{24}}}}{{1 – \tan \frac{{7\pi }}{{24}}\tan \frac{\pi }{{24}}}} }={ \tan \left( {\frac{{7\pi }}{{24}} – \frac{\pi }{{24}}} \right)\tan \left( {\frac{{7\pi }}{{24}} + \frac{\pi }{{24}}} \right) }={ \tan \frac{{6\pi }}{{24}}\tan \frac{{8\pi }}{{24}} }={ \tan \frac{\pi }{4}\tan \frac{\pi }{3} }={ 1 \cdot \sqrt 3 }={ \sqrt 3 .}$

### Example 8.

Prove that $\tan \left( {\alpha + \beta } \right) \gt \tan \alpha + \tan \beta$ if $$0 \lt \alpha \lt \large{\frac{\pi }{4}}\normalsize$$ and $$0 \lt \beta \lt \large{\frac{\pi }{4}}\normalsize.$$

Solution.

By the tangent addition formula,

${\tan \left( {\alpha + \beta } \right) }={ \frac{{\tan \alpha + \tan \beta }}{{1 – \tan \alpha \tan \beta }}.}$

The tangent function is increasing in its domain. So, if $$0 \lt \alpha \lt \large{\frac{\pi }{4}}\normalsize,$$ then $$0 \lt \tan\alpha \lt 1.$$ Similarly, if $$0 \lt \beta \lt \large{\frac{\pi }{4}}\normalsize,$$ then $$0 \lt \tan\beta \lt 1.$$

This means that the product $$\tan\alpha\tan\beta$$ is less than $$1.$$ Hence, we have

$0 \lt \tan \alpha \tan \beta \lt 1,$

$\Rightarrow – 1 \lt – \tan \alpha \tan \beta \lt 0,$

$\Rightarrow 0 \lt 1 – \tan \alpha \tan \beta \lt 1.$

We see that the denominator is positive but less than $$1.$$ Then

${\frac{{\tan \alpha + \tan \beta }}{{1 – \tan \alpha \tan \beta }} }\gt{ \tan \alpha + \tan \beta ,}$

that is,

$\tan \left( {\alpha + \beta } \right) \gt \tan \alpha + \tan \beta .$

### Example 9.

Find the tangent of the angle at which the curves $$y_1 = x^2$$ and $$y_2 = \large{\frac{1}{x}}\normalsize$$ intersect.

Solution.

The angle between two curves at the point of intersection is defined as the angle between the tangent lines to the curves at this point.

We denote the angle between the tangent lines by $$\gamma.$$ It is clear that

$\gamma = \alpha – \beta,$

where $$\alpha$$ is and $$\beta$$ are the angles that the tangent lines to the curves $$y_2=\large{\frac{1}{x}}\normalsize$$ and $$y_1=x^2$$ form with the positive $$x-$$axis.

Determine the point of intersection of the curves:

${{y_1} = {y_2},}\;\; \Rightarrow {{x^2} = \frac{1}{x},}\;\; \Rightarrow{ {x^3} = 1,}\;\; \Rightarrow {x = 1.}$

The tangent of the slope angle is equal to the value of the derivative at the point of tangency. The derivatives of the functions are given by

${y_1^\prime = \left( {{x^2}} \right)^\prime = 2x,\;\;}\kern0pt{y_2^\prime = \left( {\frac{1}{x}} \right)^\prime = – \frac{1}{{{x^2}}}.}$

Now we know the tangents of $$\alpha$$ and $$\beta:$$

${\tan \alpha = y_2^\prime\left( 1 \right) }={ – \frac{1}{{{1^2}}} }={ – 1,}$

${\tan \beta = y_1^\prime \left( 1 \right) }={ 2 \cdot 1 }={ 2.}$

The tangent of the angle between the curves is given by

${\tan \gamma = \tan \left( {\alpha – \beta } \right) }={ \frac{{\tan \alpha – \tan \beta }}{{1 + \tan \alpha \tan \beta }} }={ \frac{{ – 1 – 2}}{{1 + \left( { – 1} \right) \cdot 2}} }={ \frac{{ – 3}}{{ – 1}} }={ 3.}$

### Example 10.

Find the tangent of the angle of intersection of the curves $$y_1 = \ln x + 1$$ and $$y_2 = \large{\frac{1}{x^2}}\normalsize.$$

Solution.

The angle of intersection of two curves is the same as the angle between the tangent lines of the curves at their point of intersection.

In our case, the angle $$\gamma$$ between the tangent lines of the curves is equal to

$\gamma = \alpha – \beta ,$

where $$\alpha$$ and $$\beta$$ are the slope angles of the tangent lines.

From the other side, the tangents of $$\alpha$$ and $$\beta$$ are expressed in terms of the derivatives of the functions:

${\tan \alpha = y_2^\prime\left( {{x_0}} \right),\;\;}\kern0pt{\tan \beta = y_1^\prime\left( {{x_0}} \right),}$

where $$x_0$$ is the point of intersection.

We can easily see that the point of intersection is $$x_0 = 1,$$ since

${{y_1}\left( 1 \right) = \ln 1 + 1 }={ 0 + 1 }={ 1,}$

${{y_2}\left( 1 \right) = \frac{1}{{{1^2}}} }={ 1.}$

Compute the values of the derivatives at this point and the tangents of $$\alpha$$ and $$\beta:$$

${y_1^\prime = \left( {\ln x + 1} \right)^\prime = \frac{1}{x},}\;\; \Rightarrow {\tan\beta = y_1^\prime\left( 1 \right) = \frac{1}{1} = 1;}$

${y_2^\prime = \left( {\frac{1}{{{x^2}}}} \right)^\prime = – \frac{2}{{{x^3}}},}\;\; \Rightarrow {\tan \alpha = y_2^\prime\left( 1 \right) = – \frac{2}{{{1^3}}} = – 2.}$

To determine the angle between the curves, we use the tangent subtraction identity:

${\tan \gamma = \tan \left( {\alpha – \beta } \right) }={ \frac{{\tan \alpha – \tan \beta }}{{1 + \tan \alpha \tan \beta }} }={ \frac{{ – 2 – 1}}{{1 + \left( { – 2} \right) \cdot 1}} }={ \frac{{ – 3}}{{ – 1}} }={ 3.}$