Question

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

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

Answer 1

The correct option is B $\frac{1+\sin \theta }{1-\sin \theta }$

${(\tan \theta +\sec \theta )}^2$

$={\left(\frac{\sin \theta +1}{\cos \theta }\right)}^2$

$=\frac{1+{\sin }^2\theta +2sin\theta }{{\cos }^2\theta }$

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

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

$=\frac{1+\sin \theta }{1-\sin \theta }$.