If tanα = x+1 and tanβ = x−1, Show that 2cot(α−β) = x²

Question

If

\[ \tan\alpha=x+1 \]

and

\[ \tan\beta=x-1 \]

show that:

\[ 2\cot(\alpha-\beta)=x^2 \]

Proof

Using

\[ \tan(\alpha-\beta) = \frac{\tan\alpha-\tan\beta} {1+\tan\alpha\tan\beta} \]

\[ = \frac{(x+1)-(x-1)} {1+(x+1)(x-1)} \]

\[ = \frac{2} {1+x^2-1} \]

\[ = \frac{2}{x^2} \]

Therefore,

\[ \cot(\alpha-\beta) = \frac{x^2}{2} \]

\[ 2\cot(\alpha-\beta)=x^2 \]

Hence proved.