A) \[\sin \theta =1-\cos \theta \]
B) \[sec\theta -\tan \theta =\frac{1}{\sec \theta +\tan \theta }\]
C) \[{{\tan }^{2}}\theta -{{\sin }^{2}}\theta ={{\tan }^{2}}\theta {{\sin }^{2}}\theta \]
D) \[\sin \frac{\pi }{3}=\cos \frac{\pi }{6}\]
Correct Answer: A
Solution :
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