Question

If tan a = x +1, tanß = x - 1 show that 2 cota (alpha-B) = x^2​

Answer 1

$\large \sf \maltese \: \: \: \underline{ \underline{Question \: : }}$

If

$\bf \tan \alpha = x + 1 \: \: \: \& \: \: \: \tan \beta = x - 1$

then show 2 cot ( alpha + beta ) = x²

$\large\sf \maltese \: \: \: \underline{ \underline{Solution\: : }}$

$\bull \: \: \: \bf 2 \: \cot( \alpha - \beta ) \\ \\ \: \sf\implies 2 \ \: \frac{ \cos( \alpha - \beta )}{ \sin( \alpha - \beta )} \\ \\ \implies \sf \:2 \: \frac{ \cos \alpha \cos \beta + \sin \alpha \sin \beta }{ \sin \alpha \cos \beta - \cos \alpha \sin \beta } \times \frac{ \frac{1}{ \cos \alpha \cos \beta } }{ \frac{1}{ \cos \alpha \cos \beta } } \\ \\ \implies \sf \: 2 \: \: \frac{1 + \tan \alpha \tan \beta }{ \tan \alpha - \tan \beta } \\ \\ \implies \sf \: 2 \: \: \frac{1 + (x + 1)(x - 1)}{x + 1 - x + 1} \\ \\ \implies \sf \: \cancel{2} \frac{ \cancel{1} + {x}^{ {2}} + \cancel{ 1} }{ { \cancel{2}} } \\ \\ \\ { \large{\therefore}} { \: \: \bf \large { \underline{\boxed{ \bf{ {x}^{2} }} \: \: \: \: \: }}_{ \bigstar \star}}$