1.limx→π/21−tan(x/2)1+tan(x/2)⋅1−sinx(π−2x)3\displaystyle\lim_{x\to\pi/2}\frac{1-\tan(x/2)}{1+\tan(x/2)}\cdot\frac{1-\sin x}{(\pi-2x)^3}x→π/2lim1+tan(x/2)1−tan(x/2)⋅(π−2x)31−sinxa.1/321/321/32b.000c.1/161/161/16d.1/81/81/8Login to continueOnly logged in users canattempt or see the solution.