Question

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

Answer 1

$

\cot(a-b)=\frac{\frac{1}{(x+1)(x-1)}+1}{\frac{1}{x-1}+\frac{1}{x+1}} \\ \\ =\frac{1-1+x^2}{2} \\ \\ =\frac{x^2}{2} \\ \\ \therefore 2\cot(A-B)=x^2$

Answer 2

Answer:

See below for proof.

Step-by-step explanation:

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

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

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