Question

If tan a = x + 1 and tan ß = x - 1, then verify that: 2cot(a - B) = x²​

Answer 1

`</p><p>\cot(a-b)=((1)/((x+1)(x-1))+1)/((1)/(x-1)+(1)/(x+1)) \\ \\ =(1-1+x^2)/(2) \\ \\ =(x^2)/(2) \\ \\ \therefore 2\cot(A-B)=x^2`

Answer 2

See below for proof.

Explanation:

`\boxed{\begin{minipage}{5 cm}\underline{Trigonometric Identities}\\\\$\cot \theta=(1)/(\tan \theta)$\\\\$\tan (A \pm B)=(\tan A \pm \tan B)/(1 \mp \tan A \tan B)$\\\end{minipage}}`

`\begin{aligned}\implies 2\cot (\alpha - \beta) & =(2)/(\tan (\alpha - \beta))\\\\ & =(2)/(( \tan \alpha - \tan \beta)/(1+\tan \alpha \tan \beta))\\\\ & =(2(1+\tan \alpha \tan \beta))/( \tan \alpha - \tan \beta)\\\\ & =(2(1+(x+1)(x-1)))/( (x+1) - (x-1))\\\\& = (2(1+(x^2-x+x-1)))/(x+1-x+1)\\\\& = (2(1+(x^2-1)))/(2)\\\\& = (2x^2)/(2)\\\\& = x^2\end{aligned}`

Hence verifying that 2cot(α - β) = x².