Answer
$\lim\limits_{x \to (\pi/2)^-}~(tan~x)^{cos~x} = 1$
Work Step by Step
Let $~~~y = \lim\limits_{x \to (\pi/2)^-}~(tan~x)^{cos~x}$
$ln~y = \lim\limits_{x \to (\pi/2)^-}~ln~(tan~x)^{cos~x}$
$ln~y = \lim\limits_{x \to (\pi/2)^-}~cos~x~ln~(tan~x)$
$ln~y = \lim\limits_{x \to (\pi/2)^-}~\frac{ln~(tan~x)}{sec~x} = \frac{\infty}{\infty}$
We can use L'Hospital's Rule:
$ln~y = \lim\limits_{x \to (\pi/2)^-}~\frac{ln~(tan~x)}{sec~x}$
$ln~y = \lim\limits_{x \to (\pi/2)^-}~\frac{\frac{cos~x}{sin~x}~\cdot~\frac{1}{cos^2~x}}{\frac{sin~x}{cos^2~x}}$
$ln~y = \lim\limits_{x \to (\pi/2)^-}~\frac{cos~x}{sin^2~x}$
$ln~y=\frac{0}{1}$
$ln~y= 0$
$y = e^0$
$y = 1$