Question
If than $\alpha=\text{x}+1,\tan\beta=\text{x}-1,$ prove that $2\cot(\alpha-\beta)=\text{x}^2$

Answer

We have,
$\tan\alpha=\text{x}+1\text{ and }\tan\beta=\text{x}-1$
Now, $2\cot(\alpha-\beta)$
$=\frac{2}{\tan(\alpha-\beta)}$
$=\frac{\frac{2}{\tan\alpha-\tan\beta}}{1+\tan\alpha\tan\beta}$
$=\frac{2(1+\tan\alpha\tan\beta)}{\tan\alpha-\tan\beta}$
$=\frac{2[1+(\text{x}+1)(\text{x}-1)]}{\text{x}+1-(\text{x}-1)}$
$=\frac{2[1+\text{x}^2-1]}{\text{x}+1-\text{x}+1}$
$=\frac{2\text{x}\text{x}^2}{2}=\text{x}^2$
$\therefore2+\cot(\alpha+\beta )=\text{x}^2$
Hence proved.

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