Trigonometric Identities Question 162
Question: $ \frac{2\sin \theta ,\tan \theta (1-\tan \theta )+2\sin \theta {{\sec }^{2}}\theta }{{{(1+\tan \theta )}^{2}}}= $
[Roorkee 1975]
Options:
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
Show Answer
Answer:
Correct Answer: B
Solution:
Given expression $ =\frac{2,\sin \theta }{{{(1+\tan ,\theta )}^{2}}},{ \tan ,\theta ,(1-\tan ,\theta )+{{\sec }^{2}}\theta } $ $ =\frac{2,\sin \theta }{{{(1+\tan ,\theta )}^{2}}},{ \tan ,\theta ,-{{\tan }^{2}},\theta +1+{{\tan }^{2}}\theta } $ $ =\frac{2,\sin \theta }{1+\tan \theta } $ .
BETA
