# Mathematical Induction involving Compound Angle Formula of Tangent

Trigonometric properties and formulae can be used to perform proofs using mathematical induction. We use the following compound angle formulae for mathematical induction in this example.

\displaystyle \begin{align} \tan (\alpha + \beta) &= \frac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta} \\ \tan (\alpha-\beta) &= \frac{\tan \alpha-\tan \beta}{1 + \tan \alpha \tan \beta} \end{align}

## Part 1

Show that $\displaystyle 1 + \tan n x \tan (n+1)x = \cot x \big[\tan (n+1)x-\tan n x \big]$ using $\tan (\alpha-\beta) = \displaystyle \frac{\tan \alpha-\tan \beta}{1 + \tan \alpha \tan \beta}$.

\displaystyle \begin{align} \tan \big[nx-(n+1)x \big] &= \displaystyle \frac{\tan n x-\tan (n+1)x}{1 + \tan n x \tan (n+1)x} \\ 1 + \tan n x \tan (n+1)x &= \frac{\tan n x-\tan (n+1)x}{\tan \big[nx-(n+1)x \big]} \\ &= \frac{\tan n x-\tan (n+1)x}{\tan (-x)} \\ &= \frac{\tan n x-\tan (n+1)x}{ -\tan x} \\ &= \frac{\tan (n+1)x-\tan n x}{ \tan x} \\ &= \frac{1}{\tan x} \times \big[ \tan (n+1)x-\tan n x \big] \\ \require{AMSsymbols} \therefore 1 + \tan n x \tan (n+1)x &= \cot x \big[ \tan (n+1)x-\tan n x \big] \end{align}

## Part 2

Use mathematical induction to prove that, for all integers $n \ge 1$,

$\tan x \tan 2x + \tan 2x \tan 3x + \cdots + \tan n x \tan (n+1)x = -(n+1) + \cot x \tan (n+1)x$.

### Step 1

\begin{align} \text{For } n &= 1: \\ \text{LHS} &= 1 + \tan x \tan 2x \\ &= \cot x (\tan 2x-\tan x) \color{green}{\cdots \text{Part 1}} \\ &= \text{RHS} \\ \require{AMSsymbols} \therefore \text{The } &\text{statement is true for } n=1 \end{align}

### Step 2

$\text{Assume the statement is true for } n =k.$
$\text{That is, } \tan x \tan 2x + \tan 2x \tan 3x + \cdots + \tan k x \tan (k+1)x = -(k+1) + \cot x \tan (k+1)x$

### Step 3

$\text{Assume the statement is true for } n =k+1.$
$\text{That is, } \tan x \tan 2x + \tan 2x \tan 3x + \cdots + \tan k x \tan (k+1)x + \tan (k+1)x \tan (k+2) x = -(k+2) + \cot x \tan (k+2)x$
\begin{align} \text{LHS} &= \bbox[#F80]{\tan x \tan 2x + \tan 2x \tan 3x + \cdots + \tan k x \tan (k+1)x } + \tan (k+1)x \tan (k+2) x \\ &= \bbox[#F80]{-(k+1) + \cot x \tan (k+1)x } + \tan (k+1)x \tan (k+2) x &\color{green}{\text{Assumption}} \\ &= -(k+1) + \cot x \tan (k+1)x + \tan (k+1)x \tan (k+2) x + 1-1 \\ &= -(k+1) + \cot x \tan (k+1)x + \cot x \big[ \tan (k+2)x-\tan (k+1)x \big]-1 \\ &= -(k+2) + \bbox[#F80]{\cot x \tan (k+1)x} + \cot x \tan (k+2)x-\bbox[#F80]{\cot x \tan (k+1)x} \\ &= -(k+2) + \cot x \tan (k+2)x \\ &= \text{RHS} \end{align}
$\text{The statement is true for } n=k+1.$
$\require{AMSsymbols} \therefore \text{The statement is true for } n=1 \text{and for } n=k+1, \text{then it is true for } n=2,3, \cdots \text{ and for all integers } n \ge 1.$