A) \[{{\sin }^{2}}\alpha =1\]
B) \[{{\cos }^{2}}\alpha =1\]
C) \[{{\sin }^{2}}\alpha ={{a}^{2}}\]
D) \[{{\cos }^{2}}\alpha ={{a}^{2}}\]
Correct Answer: B
Solution :
The tangent is y \[\cos \alpha =x\sin \alpha +a\cos \alpha \] \[\Rightarrow y=x\tan \alpha +a\] It is a tangent to the circle\[{{x}^{2}}+{{y}^{2}}={{a}^{2}}\], if \[{{a}^{2}}={{a}^{2}}(1+{{\tan }^{2}}\alpha )\Rightarrow {{\sec }^{2}}\alpha =1\Rightarrow {{\cos }^{2}}\alpha =1\].
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