The set of values of x for which tan3x-tan2x-1-tan3xtan2x-1 is-

The correct option is C { nπ +π/4; n= 1, 2,3 ,.......} use>(tanA tanB/1+tanA>tanB)=tan(A B) ∴>>(tan3x tan2x/1+tan3x>tan2x)=1 & ...

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Trigonometric Ratios for Sum of Two Angles

The set of va...

Question

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

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{4}

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{n π + \frac{π}{4}; n = 1, 2, 3, . . . . . . .}

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{2 n π + \frac{π}{4}}; n = 1, 2, 3, . . . . .

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\frac{t a n 3 x - t a n 2 x}{1 + t a n 3 x t a n 2 x} = 1 \Rightarrow x = n π + \frac{π}{4}, n \in l

t a n x

x = n π + \frac{π}{2}, n \in l

x

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

n \in Z

x

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

{n π + \frac{π}{4}, n \in l}

(1 + t a n x + {tan}^{2} x)

(1 - cot x + {cot}^{2} x)

x

e^{sin x} - e^{- s ˙ i n x} \neq 4

x

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

x = n π + \frac{π}{4}, \forall n \in I

tan x

x = n π + \frac{π}{2}, n \in I

\frac{tan 3 x - tan 2 x}{1 + tan 3 x tan 2 x} = 1 i s x = n π + \frac{π}{4}

tan x

\frac{π}{2}

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