question_answer

If sec\[A=\frac{2}{\sqrt{3}}\], find the value of \[\frac{\tan \,\,A}{\cos \,\,A}+\frac{1+\sin \,\,A}{\tan \,\,A}\]

Answer:

Given, \[\sec \,A=\frac{2}{\sqrt{3}}\]

In \[\Delta \text{ }ABC\]

\[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\]

\[{{2}^{2}}={{\left( \sqrt{3} \right)}^{2}}=B{{C}^{2}}\]

\[4=3+B{{C}^{2}}\]

\[B{{C}^{2}}=4-3\]

\[B{{C}^{2}}=1\]

So,       \[\tan \,A=\frac{1}{\sqrt{3}};\cos \,A=\frac{\sqrt{3}}{2};\sin \,A=\frac{1}{2}\]

\[\frac{\tan \,A}{\cos \,A}+\frac{1+\sin \,A}{\tan \,A}=\frac{\frac{1}{\sqrt{3}}}{\frac{\sqrt{3}}{2}}+\frac{1+\frac{1}{2}}{\frac{1}{\sqrt{3}}}\]

\[=\frac{2}{3}+\frac{\frac{3}{2}}{\frac{1}{\sqrt{3}}}\]

\[=\frac{2}{3}+\frac{3\sqrt{3}}{2}\]

\[=\frac{4+9\sqrt{3}}{6}\]