Question

$\frac{{\tan }^2\Theta }{{(1+\sec \Theta )}^2}$ is equal to

• A. $\left(\frac{1-\cos \Theta }{1+\cos \Theta }\right)$
• B. $\left(\frac{1+\cos \Theta }{1-\cos \Theta }\right)$
• C. $\left(\frac{\cos \Theta -1}{\cos \Theta +1}\right)$
• D. $\left(\frac{\cos \Theta +1}{\cos \Theta -1}\right)$

Answer 1

The correct option is A $\left(\frac{1-\cos \Theta }{1+\cos \Theta }\right)$

$\frac{{\tan }^2\theta }{(1+\sec \theta )}=\frac{{\sec }^2\theta -1}{(1+\sec \theta {)}^2}$
$=\frac{(\sec \theta -1)(\sec \theta +1)}{(1+\sec \theta {)}^2}=\frac{\sec \theta -1}{1+\sec \theta }=\frac{1-\cos \theta }{1+\cos \theta }$