A) \[1\]
B) \[\frac{1}{2}\]
C) \[\frac{1}{4}\]
D) \[\frac{\sqrt{3}}{2}\]
Correct Answer: B
Solution :
| [b] We have, \[{{\tan }^{2}}A=1+2{{\tan }^{2}}B\] |
| \[\Rightarrow \,\,\,{{\sec }^{2}}A-1=1+2({{\sec }^{2}}B-1)\] \[[{{\tan }^{2}}\theta ={{\sec }^{2}}\theta =1]\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,{{\sec }^{2}}A=2+2{{\sec }^{2}}B-2\Rightarrow \,\,\,{{\sec }^{2}}A=2{{\sec }^{2}}B\] |
| \[\Rightarrow \,\,\,\,\,\,\frac{1}{{{\cos }^{2}}A}=\frac{2}{{{\cos }^{2}}B}\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\frac{{{\cos }^{2}}A}{{{\cos }^{2}}B}=\frac{1}{2}\] |
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