A) \[\frac{1+t}{1-t}\]
B) \[\frac{1-t}{1+t}\]
C) \[\frac{2t}{1-t}\]
D) \[\frac{2t}{1+t}\]
Correct Answer: A
Solution :
\[\tan 2\theta =\frac{2\tan \theta }{1-{{\tan }^{2}}\theta },\cos 2\theta =\frac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }\] \[\tan 2\theta +\sec 2\theta =\frac{2t}{1-{{t}^{2}}}+\frac{1+{{t}^{2}}}{1-{{t}^{2}}}=\frac{{{(1+t)}^{2}}}{(1-t)(1+t)}=\frac{1+t}{1-t}\].
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