Addition and Subtraction Formulas for Tangent and Cotangent

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,

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

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,

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

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:

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

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

$\cot\left( {\alpha + \beta} \right) = \frac{1 - \tan\alpha\tan\beta}{\tan\alpha + \tan\beta}$

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

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

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

$\cot\left( {\alpha - \beta} \right) = \frac{1 + \tan\alpha\tan\beta}{\tan\alpha - \tan\beta}$

See solved problems on Page 2.