Question

What is the value of $\frac{\tan ({90}^0-\theta )\sec ({180}^0-\theta )(-\theta )}{\sin ({180}^0+\theta )\cot ({360}^0-\theta )co\sec ({90}^0-\theta )}$ ?

Answer 1

The given expression is equal to
$\frac{\cot \theta (-\sec \theta )(-\sin \theta )}{(-\sin \theta )(-\cot \theta )\left(\sec \theta \right)}=1$