Question

The expression ${(\tan \Theta +sec\Theta )}^2$ is equal to

• A. $\frac{1+cos\Theta }{1-cos0}$
• B. $\frac{1+sin\Theta }{1-sin0}$
• C. $\frac{1-cos\Theta }{1+cos0}$
• D. $\frac{1-sin\Theta }{1+sin\Theta }$

Answer 1

Answer: (B)

$\frac{1+sin\Theta }{1-sin0}$

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