A) \[\frac{x}{{{x}^{2}}+1}\]
B) \[\frac{x}{{{x}^{2}}-1}\]
C) \[\frac{2x}{{{x}^{2}}-1}\]
D) \[\frac{2x}{{{x}^{2}}+1}\]
Correct Answer: C
Solution :
| [c] \[x=\sin \theta +\cos \theta \] ...(i) \[y=\sec \theta +\text{cosec}\theta \] \[=\frac{1}{\sin \theta }+\frac{1}{\cos \theta }\] \[=\frac{\sin \theta +\cos \theta }{\sin \theta \cos \theta }\] \[=\frac{x}{\sin \theta cos\theta }\] \[\because \]\[{{(sin\theta +cos\theta )}^{2}}=1+2\sin \theta \cos \theta \] \[{{x}^{2}}=1+2\sin \theta cos\theta \] \[\sin \theta \cos \theta =\frac{{{x}^{2}}-1}{2}\] [from Eq. (ii)] \[y=\frac{2x}{{{x}^{2}}-1}\] |
You need to login to perform this action.
You will be redirected in
3 sec
