| If\[\sec \theta +\tan \theta =p,\]then \[\cos \theta \] is equal to |
A) \[\frac{{{p}^{2}}+1}{{{p}^{2}}-1}\]
B) \[\frac{{{p}^{2}}-1}{{{({{p}^{2}}+1)}^{2}}}\]
C) \[\frac{2p}{{{p}^{2}}+1}\]
D) \[\frac{4{{p}^{2}}}{{{({{p}^{2}}+1)}^{2}}}\]
E) None of these
Correct Answer: C
Solution :
| \[\sec \theta +\tan \theta =p\]\[\Rightarrow \]\[\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }=p\] |
| \[\Rightarrow \] \[1+\sin \theta =p\cos \theta \] (i) |
| On squaring both sides, we get |
| \[1+{{\sin }^{2}}\theta +2\sin \theta ={{p}^{2}}{{\cos }^{2}}\theta \] |
| \[\Rightarrow \]\[1+1-{{\cos }^{2}}\theta +2\sin \theta ={{p}^{2}}{{\cos }^{2}}\theta \] |
| \[\Rightarrow \]\[2+2\sin \theta -{{\cos }^{2}}\theta ={{p}^{2}}{{\cos }^{2}}\theta \] |
| \[\Rightarrow \]\[2\,\,(1+\sin \theta )-{{\cos }^{2}}\theta ={{p}^{2}}{{\cos }^{2}}\theta \] |
| \[\Rightarrow \]\[2\times p\cos \theta -{{\cos }^{2}}\theta ={{p}^{2}}{{\cos }^{2}}\theta \][from Eq.(i)] |
| \[\Rightarrow \] \[2p\cos \theta ={{p}^{2}}{{\cos }^{2}}\theta +{{\cos }^{2}}\theta \] |
| \[\Rightarrow \] \[2p\cos \theta ={{\cos }^{2}}\theta \,\,(1+{{p}^{2}})\] |
| \[\therefore \] \[{{\cos }^{2}}\theta =\frac{2p\cos \theta }{(1+{{p}^{2}})}\] |
| \[\Rightarrow \] \[\cos \theta =\frac{2p}{1+{{p}^{2}}}\] |
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