A) \[\sqrt{\frac{l-2}{1-m}}\]
B) \[\sqrt{\frac{2-l}{1-m}}\]
C) \[\sqrt{\frac{l-2}{m-1}}\]
D) \[\sqrt{\frac{l-1}{2-m}}\]
Correct Answer: D
Solution :
| [d] \[l{{\cos }^{2}}\theta +m{{\sin }^{2}}\theta \] \[={{\cos }^{2}}\theta \frac{(cose{{c}^{2}}\theta +1)}{(cose{{c}^{2}}\theta -1)}\] \[=\frac{{{\cos }^{2}}\theta \,(1+si{{n}^{2}}\theta )}{1-\sin }\cdot \frac{{{\sin }^{2}}\theta }{{{\sin }^{2}}\theta }\] \[=\frac{{{\cos }^{2}}\theta \,(1+si{{n}^{2}}\theta )}{{{\cos }^{2}}\theta }=1+{{\sin }^{2}}\theta \] \[={{\cos }^{2}}\theta +{{\sin }^{2}}\theta +{{\sin }^{2}}\theta ={{\cos }^{2}}\theta +2{{\sin }^{2}}\theta \] \[\Rightarrow \] \[(1-1)co{{s}^{2}}\theta =(2-m)si{{n}^{2}}\theta \] \[\Rightarrow \] \[{{\tan }^{2}}\theta =\frac{l-1}{2-m}\]\[\therefore \]\[\tan \theta =\sqrt{\frac{l-1}{2-m}}\] |
You need to login to perform this action.
You will be redirected in
3 sec
