find range

2. put x= tan(t) .

y= t² + 2cost
dy/dx = 2t - 2sint =0 , or t=0
d²y/dx² >0 at t=0

hence range is [2,∞)

the above range is wrong since t- (-pi/2 , pi/2 )
hence range will be [2, f(pi/2)]

\hspace{-16}(1)\;\; $Here $\mathbf{f(x)=\frac{\sin^2x+\sin x-1}{\sin^2 x-\sin x+2}}$\\\\\\ Put $\mathbf{\sin x=t\Leftrightarrow -1\leq t \leq 1}$\\\\\\ So $\mathbf{f(t)=\frac{t^2+t-1}{t^2-t+1}}$\\\\\\ $\mathbf{\frac{d}{dt}(f(t))=\frac{-2t^2+6t+1}{(t^2-t+1)^2}}$\\\\\\ Now For Max. OR Min. $\mathbf{\frac{d}{dt}(f(t))=0}$\\\\\\ So $\mathbf{-2t^2+6t+1=0\Leftrightarrow t=\frac{-3\pm \sqrt{11}}{2}}$\\\\\\ So $\mathbf{f_{Min.}\left(\frac{3-\sqrt{11}}{2}\right)=\frac{3-2\sqrt{11}}{7}}$\\\\\\ and Max. Occur at Boundry\\\\\\ So $\mathbf{f_{Max.}\left(1\right)=\frac{1}{2}}$\\\\\\ So Range of $\mathbf{f(x)}$ is $\mathbf{\left[\frac{3-2\sqrt{11}}{7}\;,\frac{1}{2}\right]}$\\\\\\\\ $\mathbf{(2)\;\; [2,f\left(\frac{\pi}{2}\right))= \left[2,\frac{\pi^2}{4}\right)}

^ you've made a blunder :D