A) \[\frac{2}{x}\]
B) \[\frac{1}{2x}\]
C) \[\frac{{{x}^{2}}-1}{2x}\]
D) \[\frac{2x}{{{x}^{2}}-1}\]
Correct Answer: C
Solution :
| [c]\[\sec A+\tan A=x\] (1) |
| Also \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,(\sec A-\tan A)\,\,(\sec A+\tan A)=1\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,x(\sec A-\tan A)=1\] |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\sec A-\tan A=\frac{1}{x}\] .(2) |
| From eqs. (1) and (2), we get \[\tan A=\frac{{{x}^{2}}-1}{2x}\] |
You need to login to perform this action.
You will be redirected in
3 sec
