Statement – 1 : $\frac{\tan 3 x - \tan 2 x}{1 + \tan 3 x \tan 2 x} = 1, x = n π + \frac{π}{4}, n \in I$
Statement – 2 : tan x is not defined at $x = n π + \frac{π}{2}, n \in I$

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Statement-1 is True, Statement-2 is False

Statement-1 is False, Statement-2 is True

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1

answer is D.

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Detailed Solution

$∵ \frac{\tan 3 x - \tan 2 x}{1 + \tan 3 x \tan 2 x} \Rightarrow \tan (3 x - 2 x) = 1$

Or tan x = 1 Or $x = n π + \frac{π}{4}, n \in I$

But at this value, tan 2x is undefined, hence there is no solution and $\tan x = \frac{\sin x}{\cos x}$

$∴ \tan x$ is not defined when cos x = 0 Or $x = n π + \frac{π}{2}, n \in I$

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