A) \[-1\]
B) \[1\]
C) \[2\]
D) \[\frac{1}{2}\]
Correct Answer: B
Solution :
| [b]\[\sqrt{(1-{{\cos }^{2}}\theta )\,{{\sec }^{2}}\theta }=\sqrt{{{\sin }^{2}}\theta \cdot {{\sec }^{2}}\theta }\] \[[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1]\] |
| \[=\sqrt{{{\sin }^{2}}\theta \times \frac{1}{{{\cos }^{2}}\theta }}=\sqrt{{{\tan }^{2}}\theta }=\tan \theta \] |
| \[=k\tan \theta \] |
| On comparing, we get |
| \[k=1\] |
You need to login to perform this action.
You will be redirected in
3 sec
