Question

tan(π/4 + x) / tan(π/4 - x) = {( 1+ tan x )/ (1- tan x )}^2 ​

Answer 1

$\bf \underline {\underline{Solution-}}$

$\bf \: Used \: Formulas -$

$\boxed{ \bf \:1. \: \: tan \:(A + B ) = \frac{tanA \: + tanB}{1 -tanA.tanB } }$

$\bf \: LHS = \frac{ tan( \frac{\pi}{4} + x)}{tan (\frac{\pi}{4} - x)}$

$\bf \: let \: First \: we \: solve \: \because \: tan( \frac{\pi}{4} + x) \\ \\ \\ \implies \bf \: \frac{tan \frac{\pi}{4} + tan \: x}{1 - tan \frac{\pi}{4} .tan \: x} \\ \\ \\ \implies \: \bf \: we \: can \: write \: it \: as \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{1 + tan \: x}{1 - tan \: x} ........(1) \: \: \: \: \: \: \: \: \huge \{ \tiny \: tan(a + b) = \frac{tan \: a + tan \: b}{1 - tan \: a. \: tan \: b} \\ \\ \\ \implies \bf \: like \: that \: tan( \frac{\pi}{4} - x) = \frac{1 - tan \: x}{1 + tan \: x} .........(2) \\ \\ \\ \bf \: now \:dividing \: the \: equation(1) \: to \: equation(2)\: \\ \\ \\ \therefore \bf \: LHS \: = \frac{tan( \frac{\pi}{4} + x)}{tan( \frac{\pi}{4} - x)} = \frac{ \frac{1 + tan \: x}{1 - tan \: x} }{ \frac{1 - tan \: x}{1 + tan \: x} } \\ \\ \\ \implies \bf \: \frac{1 + tan \: x}{1 - tan \: x} \times \frac{1 + tan \: x}{1 - tan \: x} \\ \\ \\ \implies \bf \: {( \frac{1 + tan \: x}{1 - tan \: x} )}^{2}$