A) \[\frac{1}{2m}\]
B) \[m\]
C) \[2\,\,m\]
D) \[\frac{m}{2}\]
Correct Answer: C
Solution :
| \[\sin \theta +\cos \theta =m\] |
| and \[\sec \theta +\text{cosec}\theta \text{=n}\] |
| \[\Rightarrow \] \[\frac{\sin \theta +\cos \theta }{\sin \theta \cos \theta }=n\] |
| \[\Rightarrow \] \[\frac{m}{\sin \theta \cos \theta }=n\] [from Eq. (i)] |
| \[\Rightarrow \] \[\frac{1}{\sin \theta \cos \theta }=\frac{n}{m}\]or\[\sin \theta \cos \theta =\frac{m}{n}\] Squaring Eq. (i) |
| \[(si{{n}^{2}}\theta +co{{s}^{2}}\theta )+2sin\theta cos\theta ={{m}^{2}};1+2\frac{m}{n}={{m}^{2}}\] |
| \[\Rightarrow \] \[\frac{2\,\,m}{n}={{\text{m}}^{2}}-1\]\[\Rightarrow \]\[2m=({{\text{m}}^{2}}-1)n\] |
You need to login to perform this action.
You will be redirected in
3 sec
