The set of values of x for which tan 3 x -tan 2 x -1-tan 3 x -tan 2 x -1 is where n - Z A- n -4B- n -6C- n -3D- -

The correct option is D ϕWe have, tan3x tan2x1+tan 3x.·tan2x = 1 ⇒tan(3x 2x) = 1 ⇒tan x = 1 ⇒tan x = tanπ4 ⇒ x = nπ + π4, n∈ℤ But for this value of x, tan 2 ...

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Standard XII

Mathematics

General Solution of Trigonometric Equation

The set of va...

Question

The set of values of $x$ for which $\frac{tan 3 x - tan 2 x}{1 + tan 3 x \cdot tan 2 x} = 1$ is (where $n \in Z$ )

{n π + \frac{π}{4}}

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{n π + \frac{π}{6}}

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{n π + \frac{π}{3}}

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ϕ

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Solution

The correct option is D $ϕ$
We have,
$\frac{tan 3 x - tan 2 x}{1 + tan 3 x . \cdot tan 2 x} = 1 \Rightarrow tan (3 x - 2 x) = 1 \Rightarrow tan x = 1$
$\Rightarrow tan x = tan \frac{π}{4} \Rightarrow x = n π + \frac{π}{4}, n \in Z$
But for this value of $x$ , tan $2 x$ is not defined.
Hence the solution set for $x$ is $ϕ$ .

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The set of values of x for which tan3x−tan2x1+tan3x⋅tan2x=1 is (where n∈Z)

The correct option is D ϕWe have, tan3x−tan2x1+tan3x.⋅tan2x=1⇒tan(3x−2x)=1⇒tanx=1 ⇒tanx=tanπ4⇒x=nπ+π4,n∈Z But for this value of x, tan 2x is not defined. Hence the solution set for x is ϕ.

The set of values of $x$ for which $\frac{tan 3 x - tan 2 x}{1 + tan 3 x \cdot tan 2 x} = 1$ is (where $n \in Z$ )