A) 1
B) 2
C) -1
D) 0
Correct Answer: C
Solution :
(c): \[\frac{{{\sin }^{2}}\theta -2{{\sin }^{4}}\theta }{2{{\cos }^{4}}\theta -{{\cos }^{2}}\theta }-{{\sec }^{2}}\theta \] \[\frac{{{\sin }^{2}}\left( 1-2{{\sin }^{2}}\theta \right)}{2{{\cos }^{2}}\theta \left( 2{{\cos }^{2}}\theta -1 \right)}-{{\sec }^{2}}\theta \] \[\frac{{{\sin }^{2}}\left( 1-2\left( 1-{{\cos }^{2}}\theta \right) \right)}{{{\cos }^{2}}\theta \left( 2{{\cos }^{2}}\theta -1 \right)}-{{\sec }^{2}}\theta \] \[={{\tan }^{2}}\theta \frac{\left( 2{{\cos }^{2}}\theta -1 \right)}{\left( 2{{\cos }^{2}}\theta -1 \right)}-{{\sec }^{2}}\theta \] \[=ta{{n}^{2}}\theta -se{{c}^{2}}\theta =-[se{{c}^{2}}\theta -tan\theta ]=-1\]
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