A) \[\frac{\sin \,\theta }{1+\tan \theta }\]
B) \[\frac{2\,\sin \theta }{1+\tan \theta }\]
C) \[\frac{2\sin \theta }{{{(1+\tan \theta )}^{2}}}\]
D) None of these
Correct Answer: B
Solution :
Given expression \[=\frac{2\,\sin \theta }{{{(1+\tan \,\theta )}^{2}}}\,\left\{ \tan \,\theta \,(1-\tan \,\theta )+{{\sec }^{2}}\theta \right\}\] \[=\frac{2\,\sin \theta }{{{(1+\tan \,\theta )}^{2}}}\,\left\{ \tan \,\theta \,-{{\tan }^{2}}\,\theta +1+{{\tan }^{2}}\theta \right\}\]\[=\frac{2\,\sin \theta }{1+\tan \theta }\].
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