A) \[\cos \theta \]
B) \[\sin \theta \]
C) \[\sec \theta \]
D) \[\cos \theta \]
E) \[\tan \theta \]
Correct Answer: A
Solution :
\[\therefore \] \[\frac{1-{{t}^{2}}}{1+{{t}^{2}}}=\frac{1-{{\tan }^{2}}(\theta /2)}{1+{{\tan }^{2}}(\theta /2)}\] \[\left[ \because t=\tan \frac{\theta }{2}given \right]\] \[=\cos \theta \]
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