Question

$(\tan \theta \sin \theta +\cos \theta {)}^2-{\left(\frac{\sec }{cosec}\right)}^2=$

[1 mark]

• A. 0
• B. 2
• C. -1
• D. 1

Answer 1

The correct option is D 1

$(\tan \theta \sin \theta +\cos \theta {)}^2-{\left(\frac{\sec }{cosec}\right)}^2$

$=\left(\frac{\sin \theta }{\cos \theta }+\cos \theta \right)-{\left(\frac{\sin \theta }{\cos \theta }\right)}^2$

$={\left(\frac{{\sin }^2\theta +{\cos }^2\theta }{\cos \theta }\right)}^2-{\tan }^2\theta$

$={\left(\frac{1}{\cos \theta }\right)}^2-{\tan }^2\theta$

$={\sec }^2\theta -{\tan }^2\theta$

$=1$