Question

The solution set of the equation

$\tan (4k+2)x-\tan (4k+1)x-\tan (4k+2)x\cdot \tan (4k+1)x=1;k\in I$ is

• A. $\varphi$
• B. $\frac{\pi }{4}$
• C. $\{n\pi +\frac{\pi }{4},n\in I\}$
• D. $\{2n\pi +\frac{\pi }{4},n\in I\}$

Answer 1

Answer: (A)

$\varphi$

Domain of the given equation is

$x\in R-\{\frac{\pi }{4},\frac{(2k+1)\pi }{(4k+1)2}\}$

Now, the equation can be written as

$\frac{\tan (4k+2)x-\tan (4k+1)x}{1+\tan (4k+2)x\cdot \tan (4k+1)x}=1$

$\Rightarrow \tan \left(x\right)=1$

$\left(\text{Using}\tan \right(A-B)=\frac{\tan A-\tan B}{1+\tan A\cdot \tan B})$

Now, solving for $\tan x=1$, we get

$x=n\pi +\frac{\pi }{4}$
But this value is not in domain.

$\therefore$ their is no solution

Hence, option A.