The set of values of x for which tan 3x - tan 2x1- tan 3x tan 2x -1 is

Tan 3x tan 2x1+ tan 3x nbsp; tan 2x =1 ⇒tan (3x 2x) =1 ⇒tan x =1 ⇒ x = nbsp;n π + π4 but for this value of x, tan 2x is not defined ⇒ Hence, solution set ...

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Standard XI

Mathematics

General Solution of Trigonometric Equation

The set of va...

Question

The set of values of $x$ for which $\frac{tan 3 x - tan 2 x}{1 + tan 3 x tan 2 x} = 1$ is

n π + \frac{π}{4}, n \in Z

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ϕ

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n π \pm \frac{π}{4}, n \in Z

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2 n π + \frac{π}{4}, n \in Z

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Solution

The correct option is B $ϕ$
$\frac{tan 3 x - tan 2 x}{1 + tan 3 x tan 2 x} = 1 \Rightarrow tan (3 x - 2 x) = 1 \Rightarrow tan x = 1 \Rightarrow x = n π + \frac{π}{4}$
but for this value of $x$ , $tan 2 x$ is not defined
$\Rightarrow$ Hence, solution set $= ϕ$

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The set of values of x for which tan3x−tan2x1+tan3xtan2x=1 is

The correct option is B ϕtan3x−tan2x1+tan3xtan2x=1⇒tan(3x−2x)=1⇒tanx=1⇒x=nπ+π4 but for this value of x, tan2x is not defined ⇒ Hence, solution set =ϕ

The set of values of $x$ for which $\frac{tan 3 x - tan 2 x}{1 + tan 3 x tan 2 x} = 1$ is

The correct option is B $ϕ$
$\frac{tan 3 x - tan 2 x}{1 + tan 3 x tan 2 x} = 1 \Rightarrow tan (3 x - 2 x) = 1 \Rightarrow tan x = 1 \Rightarrow x = n π + \frac{π}{4}$
but for this value of $x$ , $tan 2 x$ is not defined
$\Rightarrow$ Hence, solution set $= ϕ$