A) \[\frac{{{a}^{2}}+1}{2a}\]
B) \[\frac{2a}{{{a}^{2}}+1}\]
C) \[\frac{{{a}^{2}}-1}{2a}\]
D) \[\frac{2a}{{{a}^{2}}-1}\]
Correct Answer: B
Solution :
[b] Given, \[\sec A+\tan A=a\]  \[\Rightarrow \] \[{{\sec }^{2}}A+{{\tan }^{2}}A+2\sec A\cdot \tan \,A={{a}^{2}}\] (On squaring both parts) \[\Rightarrow \] \[1+{{\tan }^{2}}A+{{\tan }^{2}}A+2\sec \,A\cdot tan\,A={{a}^{2}}\] \[\Rightarrow \] \[1+2{{\tan }^{2}}A+2\sec A\cdot \tan A={{a}^{2}}\] \[\Rightarrow \] \[2\tan \,A\,(tan\,A+sec\,A)={{a}^{2}}-1\] \[\Rightarrow \] \[2\tan A\cdot a=({{a}^{2}}-1)\]\[\Rightarrow \]\[\tan A=\frac{{{a}^{2}}-1}{2a}\] Now, in \[\Delta ABC\,{{(AC)}^{2}}={{(AB)}^{2}}+{{(BC)}^{2}}\] \[{{(AC)}^{2}}={{(2a)}^{2}}+{{({{a}^{2}}-1)}^{2}}\] \[{{(AC)}^{2}}=4{{a}^{2}}+{{a}^{4}}+1-2{{a}^{2}}={{a}^{4}}+1+2{{a}^{2}}\]\[{{(AC)}^{2}}={{({{a}^{2}}+1)}^{2}}\]\[\Rightarrow \]\[AC=({{a}^{2}}+1)\] \[\therefore \]\[\cos A=\frac{AB}{AC}\cos A=\frac{2a}{{{a}^{2}}+1}\] |
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