A) \[x-1\]
B) \[\frac{1}{x}\]
C) \[\frac{1}{x+1}\]
D) \[\frac{1}{1-x}\]
Correct Answer: B
Solution :
\[x=\frac{\cos \theta }{1-\sin \theta }=\frac{\cos \theta (1+\sin \theta )}{(1-\sin \theta )(1+sin\theta )}\] \[=\frac{\cos \theta (1+\sin \theta )}{1-{{\sin }^{2}}\theta }\] \[=\frac{1+\sin \theta }{\cos \theta }\] \[\therefore \,\,\,\frac{\cos \theta }{1+\sin \theta }=\frac{1}{x}\]
You need to login to perform this action.
You will be redirected in
3 sec
