Question

Solve:
$\frac{\cos (90+\theta )\sec (-\theta )\tan ({180}^{\circ }-\theta )}{\sec ({360}^{\circ }-\theta )\sin ({180}^{\circ }+\theta )\cot ({90}^{\circ }-\theta )}=$ ?

Answer 1

$\cos (90+\theta )=-\sin \theta$

$\sec (-\theta )=\sec \theta$

$\tan (180-\theta )=-\tan \theta$

$\sec (360-\theta )=\sec \theta$

$\sin (180+\theta )=-\sin \theta$

$\cot (90-\theta )=\tan \theta$

$\frac{\cos (90+\theta )\sec (-\theta )\tan (180-\theta )}{\sec (360-\theta )\sin (180+\theta )\cot (90-\theta )}$

$=\frac{(-\sin \theta )\left(\sec \theta \right)(-\tan \theta )}{\left(\sec \theta \right)(-\sin \theta )\left(\tan \theta \right)}$

$=-1$