The general solutions of $\sin ^{2} x \cdot \sec x=\tan x-\sin x+1$ is

The general solutions of $\sin ^{2} x \cdot \sec x=\tan x-\sin x+1$ is
  1. $x=n \pi+(-1)^{n} \frac{\pi}{4}$ or $x=m \pi+\frac{3 \pi}{4} ; m, n \in z$
  2. $x=n \pi+(-1)^{n} \frac{\pi}{2} \quad$ or $\quad x=m \pi+\frac{3 \pi}{4} ; \quad m, n \in z$
  3. $x=n \pi+(-1)^{n} \frac{\pi}{2} \quad$ or $\quad x=m \pi+\frac{5 \pi}{4} ; m, n \in z$
  4. $x=n \pi+(-1)^{n} \frac{\pi}{4}$ or $x=m \pi+\frac{5 \pi}{4} ; m, n \in z$

Solution

$\sin ^{2} x \sec x=\tan x-\sin x+1$ $\therefore \quad \frac{\sin x \cdot \sin x}{\cos x}=\frac{\sin x}{\cos x}-\sin x+1$ $\therefore \quad \sin x \cdot \sin x=\sin x-\sin x \cos x+\cos x$ $\therefore \sin x(\sin x+\cos x)=\sin x+\cos x$ $\therefore \sin x(\sin x+\cos x)-(\sin x+\cos x)=0$ $\therefore(\sin x+\cos x)(\sin x-1)=0$ $\therefore \sin x+\cos x=0$ or $\sin x=1$ $\therefore \tan x=-1$ $\therefore x=\sin \pi+\frac{3 \pi}{4}$ or $x=n \pi+(-1)^{n} \frac{\pi}{2} \ldots \mathrm{m}, \mathrm{n} \in \mathrm{Z}$

Asked in: MHT CET 2020 (19 Oct Shift 1)

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