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) \[\frac{1-sin\theta }{1+sin\theta }\]
Correct Answer: B
Solution :
\[{{(\tan \theta +\sec \theta )}^{2}}={{\left( \frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta } \right)}^{2}}\] \[=\frac{{{(1+\sin \theta )}^{2}}}{{{\cos }^{2}}\theta }\] \[=\frac{{{(1+\sin \theta )}^{2}}}{1-{{\sin }^{2}}\theta }\] \[=\frac{{{(1+\sin \theta )}^{2}}}{(1-\sin \theta )\,(1+\sin \theta )}\] \[=\frac{1+\sin \theta }{1-\sin \theta }\]
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