As \(x\to\pi/2\): \(\cos x\to0^-\) and \(\tan x\to-\infty\). Let \(L=\lim(1+\cos x)^{\tan x}\). Take \(\ln\): \(\ln L=\lim\tan x\cdot\ln(1+\cos x)\). As \(x\to\pi/2\), let \(u=x-\pi/2\to0\): \(\cos x=-\sin u\approx-u\) and \(\tan x=\cot(-u)\approx-1/u\). So \(\ln L\approx(-1/u)\ln(1-u)\approx(-1/u)(-u)=1\). Thus \(L=e^1=e\). Wait: \(\ln(1+\cos x)\approx\ln(1-u)\approx-u\); \(\tan x\approx-\frac{1}{u}\). Product: \((-\frac{1}{u})(-u)=1\). So \(L=e^1=e\). Hmm but let me check sign: as \(x\to(\pi/2)^-\), \(\tan x\to+\infty\) and \(\cos x\to0^+\). \(\tan x\cdot\ln(1+\cos x)\approx(+\infty)\cdot\ln(1)\to\) indeterminate. Using L'Hopital: \(\to0\cdot\infty\). Limit \(=1\Rightarrow L=e\). But also as \(x\to(\pi/2)^+\): \(\tan x\to-\infty\), \(\cos x\to0^-\), base becomes negative — not real. So limit from left: \(L=e\). But answer (A) says 1 and answer (B) says e — this is option (B)=e... re-indexing: option 0=(A)=1, option 1=(B)=e. Correct answer is \(e\) = index 1.