Question

$\frac{\cos (90°+\theta )\text{cosec}(270°+\theta )\cos (180°+\theta )}{\sec (-\theta )\sin (270°+\theta )\sin (360°-\theta )}=$

• A. $-1$
• B. $1$
• C. ${\cot }^2\theta$
• D. $\tan \theta$

Answer 1

The correct option is A $-1$

$\frac{\cos (90°+\theta )\text{cosec}(270°+\theta )\cos (180°+\theta )}{\sec (-\theta )\sin (270°+\theta )\sin (360°-\theta )}$

$=\frac{-\sin \theta (-\sec \theta )(-\cos \theta )}{\sec \theta (-\cos \theta )(-\sin \theta )}=-1$