Question

f(x)=$\frac{1}{sinx+1}$, Range=

Answer 1

$f\left(x\right)=\frac{1}{1+\sin x}$

$y=\frac{1}{1+\sin x}$

$y+y\sin x=1$

$\sin x=\frac{1-y}{g}$

$-1\le \frac{1-y}{y}\le 1$

$\frac{1-y}{y}-1\ge 0$

$\frac{1-y}{y}+1\ge 0$

$\Rightarrow \frac{1-2y}{y}\le 0$ & $\frac{1}{y}\ge 0$

$\Rightarrow \frac{1-2y}{y}\le 0$ & $\frac{1}{y}\ge 0$

$y\in (-\infty ,0)\cup (\frac{1}{2},\infty )$ $y\in (0,\infty )$

$\therefore y\in R-0$.