A) \[1/\sqrt{11}\]
B) \[-1/\sqrt{11}\]
C) \[\sqrt{\frac{10}{11}}\]
D) \[-\sqrt{\frac{10}{11}}\]
Correct Answer: C
Solution :
We have \[\tan \theta =-\frac{1}{\sqrt{10}},\]therefore \[\theta \]is in IV quadrant. So \[\cos \theta =+ve\]. Now \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \Rightarrow 1+\frac{1}{10}={{\sec }^{2}}\theta \] \[\Rightarrow {{\sec }^{2}}\theta =\frac{11}{10}\Rightarrow \cos \theta =\sqrt{\left( \frac{10}{11} \right)}\].
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