Daily Question 111

Daily Question 111

$\textbf{Q111.}$ The set of values of $x$ for which $\dfrac{\tan 3x - \tan 2x}{1+\tan 3x\tan2x}=1$

(A) $x=n\pi+\dfrac{\pi}{4}$

(B) $x=n\pi\pm\dfrac{\pi}{4}$

(C) $x=n\pi-\dfrac{\pi}{4}$

(D) none of these

$\textbf{Ans.} (D)$

$\textbf{Sol.}$ Given $\dfrac{\tan 3x - \tan 2x}{1+\tan 3x\tan2x}=1$

$\implies \tan({3x-2x})=1$

$\implies \tan x=1=\tan\dfrac{\pi}{4}$

$\implies x=n\pi+\dfrac{\pi}{4}$

But $\tan 2x=\tan\left(2\left(n\pi+\dfrac{\pi}{4}\right)\right)=$ $\tan\left(2n\pi+\dfrac{\pi}{2}\right)=$ $\tan\dfrac{\pi}{2}=\infty$

Which does not satisfy the given equation. So, no solution exits.

Daily Question 110

Compiled Daily Questions 1-100