A) -1
B) 0
C) 1
D) none of these
Correct Answer: B
Solution :
[b] \[{{\tan }^{2}}\theta =2{{\tan }^{2}}\phi +1\] Or \[1+{{\tan }^{2}}\theta =2(1+ta{{n}^{2}}\phi )\] \[\Rightarrow \,{{\sec }^{2}}\theta =2{{\sec }^{2}}\phi \] \[\Rightarrow \,{{\cos }^{2}}\phi =2{{\cos }^{2}}\theta \] \[=1+\cos 2\theta \] \[\Rightarrow \cos 2\theta ={{\cos }^{2}}\phi -1\] \[=-{{\sin }^{2}}\phi \] \[\Rightarrow {{\sin }^{2}}\phi +\cos 2\theta =0\]
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