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