Answer:
See explanation
Step-by-step explanation:
[tex]tan \: A = \frac{2 \sqrt{a} }{a - 1} ...(given) \\ \because \: {sec}^{2} \: A = 1 + {tan}^{2}\: A \\ \therefore\: {sec}^{2} \: A = 1 + \bigg(\frac{2 \sqrt{a} }{a - 1} \bigg)^{2} \\\\ \hspace{38 pt} = 1 + \frac{4a}{ {(a - 1)}^{2} } \\ \\ \hspace{38 pt}= \frac{(a - 1)^{2} + 4a}{ {(a - 1)}^{2} } \\ \\\hspace{38 pt} = \frac{a^{2} - 2a + 1+ 4a}{ {(a - 1)}^{2} } \\ \\ \hspace{38 pt}= \frac{a^{2} + 2a+ 1}{ {(a - 1)}^{2} } \\ \\ \hspace{38 pt}= \frac{(a + 1)^{2}}{ {(a - 1)}^{2} } \\ \\ \therefore {sec}^{2} \: A = \bigg(\frac{a + 1}{ {a - 1}} \bigg) ^{2} \\ \\ \therefore \: {sec} \: A = \pm \bigg(\frac{a + 1}{ {a - 1}} \bigg)\\ \\ [/tex]
In the question It is not mentioned that in which quadrant does angle A lie, so we will assume it to be in first quadrant.
[tex] \therefore \: {sec} \: A = \bigg(\frac{a + 1}{ {a - 1}} \bigg)\\ \\
\red{ \boxed{ \bold{\therefore \: {cos} \: A = \bigg(\frac{a - 1}{ {a + 1}} \bigg)}}} \\ \\ {sin} \: A ={cos} \: A \times {tan} \: A \\ \\ \hspace{25 pt}=\bigg(\frac{a - 1}{ {a + 1}} \bigg) \times \frac{2 \sqrt{a} }{a - 1} \\ \\ \purple {\boxed { \bold{{sin} \: A = \frac{2 \sqrt{a} }{a + 1}}}}[/tex]
