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