A) 2
B) 3
C) \[\frac{2328}{745}\]
D) \[\frac{2328}{845}\]
Correct Answer: D
Solution :
\[\tan \phi =2\frac{2}{5}=\frac{12}{5}\] \[\therefore \,\,\,{{\tan }^{2}}\phi ={{\left( \frac{12}{5} \right)}^{2}}=\frac{144}{25}\] \[\Rightarrow \,\,\,{{\tan }^{2}}\phi +1=\frac{144}{25}+1\] \[\Rightarrow \,\,{{\sec }^{2}}\phi =\frac{144+25}{25}=\frac{169}{25}\] \[\therefore \,\,\,\,\,\,\sec \phi =\sqrt{\frac{169}{25}}=\frac{13}{5}\] \[\cos \phi =\frac{1}{\sec \phi }=\frac{5}{13}\] \[\Rightarrow \,\,\,\,{{\cos }^{2}}\phi ={{\left( \frac{5}{13} \right)}^{2}}=\frac{25}{169}\] \[\Rightarrow \,\,\,1-{{\cos }^{2}}\theta =1-\frac{25}{169}\] \[=\frac{169-25}{169}=\frac{144}{169}\] \[\therefore \,\,\,\sin \phi =\sqrt{\frac{144}{169}}=\frac{12}{13}\] \[\sin \phi .\cos \phi +\tan \phi \] \[=\frac{12}{13}\times \frac{5}{13}+\frac{12}{5}=\frac{60}{169}+\frac{12}{5}\] \[=\frac{60\times 5+12\times 169}{845}=\frac{300+2028}{845}\] \[=\underline{\frac{\mathbf{2328}}{\mathbf{845}}}\]
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