Question

The set of values of x satisfying the equation $\frac{\tan 3x-\tan 2x}{1+\tan 3x\tan 2x}=1$ is______________.

Answer 1

$\frac{\tan 3x-\tan 2x}{1+\tan 3x\tan 2x}=1$

i.e tan (3x − 2x) = 1 ( using identity :- tan (x − y) = $\frac{\tan x-\tan y}{1+\tan x\tan y}$)
i.e tan x = 1
i.e x = tan⁻¹ $\frac{\mathrm{\pi }}{4}$

​​i.e x = $n\mathrm{\pi }+\frac{\mathrm{\pi }}{4}$ ; n ∈ Z