A) \[\frac{{{x}^{2}}+1}{x}\]
B) \[\frac{{{x}^{2}}+1}{2x}\]
C) \[\frac{{{x}^{2}}-1}{2x}\]
D) \[\frac{{{x}^{2}}-1}{x}\]
Correct Answer: B
Solution :
We have, \[\sec \theta +\tan \theta =x\] \[\Rightarrow \] \[\left( \frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta } \right)=x\] \[\Rightarrow \] \[1+\sin \theta =x\cos \theta \] ?(i) Now, squaring on both sides \[{{(1+\sin \theta )}^{2}}={{(x\,\cos \theta )}^{2}}\] \[\Rightarrow \] \[1+{{\sin }^{2}}\theta +2\sin \theta ={{x}^{2}}{{\cos }^{2}}\theta \] \[\Rightarrow \] \[1+1-{{\cos }^{2}}\theta +2\sin \theta ={{x}^{2}}{{\cos }^{2}}\theta \] \[[\therefore \,\,\,{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1]\] \[\Rightarrow \] \[2(1+\sin \theta )=co{{s}^{2}}\theta ({{x}^{2}}+1)\] \[\Rightarrow \]\[\frac{1}{{{\cos }^{2}}\theta }=\frac{{{x}^{2}}+1}{2(1+\sin \theta )}\,\,\Rightarrow \frac{1}{{{\cos }^{2}}\theta }=\frac{{{x}^{2}}+1}{2(x\cos \theta )}\] [using (i)] \[\Rightarrow \] \[\sec \theta =\frac{{{x}^{2}}+1}{2x}\] [\[\therefore \] \[\frac{1}{\cos \theta }=\sec \theta \]]
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