Q117 / 180 · COMEDK 2012 · Mathematics

The general solution of $\tan x-\sin x=1-\tan x \cdot \sin x$

  • Correct answer
    A
    $x=n \pi+\frac{\pi}{4}$ and or $x=n \pi+(-1)^{n}\left(\frac{-\pi}{2}\right)$
  • B
    $x=\frac{n \pi}{4}-\frac{\pi}{4}$ and or $x=n \pi+(-1)^{n}\left(-\frac{\pi}{2}\right)$
  • C
    $x=n \pi+\frac{\pi}{4}$
  • D
    $x=n \pi+\frac{\pi}{6}$ and or $\pi=n \pi+(-1)^{n}\left(-\frac{\pi}{2}\right)$

Solution

$\tan x-\sin x=1-\tan x \sin x$ $\Rightarrow \tan x+\tan x \sin x=1+\sin x$ $\tan x(1+\sin x)-(1+\sin x)=0$ $(1+\sin x)(\tan x-1)=0$ $\Rightarrow \quad \tan x-1=0$ or $1+\sin x=0$ $\tan x=1=\tan \frac{\pi}{4} \quad$ or $\sin x=-1=\sin \left(-\frac{\pi}{2}\right)$ $x=n \pi+\frac{\pi}{4} \quad$ or $\quad x=n \pi+(-1)^{n}\left(-\frac{\pi}{2}\right)$

Hence, the correct answer is option: 'A'.

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