Answer
$$1$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \pi /{2^ - }} {\left( {\sin x} \right)^{\tan x}} \cr
& {\text{Evaluating}} \cr
& \mathop {\lim }\limits_{x \to \pi /{2^ - }} {\left( {\sin x} \right)^{\tan x}} = {\left( {\sin \frac{\pi }{2}} \right)^{\tan \pi /2}} = {1^\infty } \cr
& {\text{This limit has the form }}{1^\infty }.{\text{ }}\left( {{\text{Using theorem 2}}{\text{.12}}} \right) \cr
& {\text{Nothing that }}{\left( {\sin x} \right)^{\tan x}} = {e^{\tan x\ln \sin x}},{\text{ then}} \cr
& \mathop {\lim }\limits_{x \to \pi /{2^ - }} {\left( {\sin x} \right)^{\tan x}} = \mathop {\lim }\limits_{x \to {0^ + }} {e^{\tan x\ln \sin x}} = {e^{\mathop {\lim }\limits_{x \to \pi /{2^ - }} \left( {\tan x\ln \sin x} \right)}} \cr
& {\text{The first step is to evaluate }} \cr
& L = \mathop {\lim }\limits_{x \to \pi /{2^ - }} \left( {\tan x\ln \sin x} \right) = \mathop {\lim }\limits_{x \to \pi /{2^ - }} \frac{{\ln \sin x}}{{\frac{1}{{\tan x}}}} = \mathop {\lim }\limits_{x \to \pi /{2^ - }} \frac{{\ln \sin x}}{{\cot x}} \cr
& L = \mathop {\lim }\limits_{x \to \pi /{2^ - }} \frac{{\ln \sin x}}{{\cot x}} = \frac{0}{0} \cr
& {\text{Using the L'Hopital's rule}} \cr
& L = \mathop {\lim }\limits_{x \to \pi /{2^ - }} \frac{{\ln \sin x}}{{\cot x}} = \mathop {\lim }\limits_{x \to \pi /{2^ - }} \frac{{\cos x}}{{ - {{\csc }^2}x}} \cr
& \mathop {\lim }\limits_{x \to \pi /{2^ - }} \frac{{\cos x}}{{ - {{\csc }^2}x}} = \frac{{\cos \left( {\pi /2} \right)}}{{ - {{\csc }^2}\left( {\pi /2} \right)}} = \frac{0}{{ - 1}} = 0 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to \pi /{2^ - }} {\left( {\sin x} \right)^{\tan x}} = \mathop {\lim }\limits_{x \to {0^ + }} {e^{\tan x\ln \sin x}} = {e^{\mathop {\lim }\limits_{x \to \pi /{2^ - }} \left( {\tan x\ln \sin x} \right)}} = {e^0} = 1 \cr
& \mathop {\lim }\limits_{x \to \pi /{2^ - }} {\left( {\sin x} \right)^{\tan x}} = 1 \cr} $$
