Answer
$ 2$
Work Step by Step
$\frac{(sin\theta+cos\theta)(sin\theta+cos\theta)-1}{sin\theta cos\theta}
=\frac{sin^2\theta+cos^2\theta+2sin\theta cos\theta-1}{sin\theta cos\theta}
=\frac{1+2sin\theta cos\theta-1}{sin\theta cos\theta}
=\frac{2sin\theta cos\theta}{sin\theta cos\theta}
=2$
