Question

Simplify the following expression:
$\frac{\tan ({180}^{\circ }-\alpha )\cos ({180}^{\circ }-\alpha )\tan ({90}^{\circ }-\alpha )}{\sin ({90}^{\circ }+\alpha )\cot ({90}^{\circ }-\alpha )\tan ({90}^{\circ }+\alpha )}.$

Answer 1

$\frac{\tan (180°-\alpha )\cos (180°-\alpha )\tan (90°-\alpha )}{\sin (90°-\alpha )\cot (90°-\alpha )\tan (90°+\alpha )}$

$=\frac{(-\tan \alpha )\times (-\cos \alpha )\times \cot \alpha }{\cos \alpha \times tan\alpha \times (-cot\alpha )}$

$=-1$