Question

(If tan A=)
PLZ ANSWER IT FAST...​

Answer 1

See explanation

Explanation:

`tan \: A = (2 √(a) )/(a - 1) ...(given) \\ \because \: {sec}^(2) \: A = 1 + {tan}^(2)\: A \\ \therefore\: {sec}^(2) \: A = 1 + \bigg((2 √(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)\\ \\`

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.

`\therefore \: {sec} \: A = \bigg(\frac{a + 1}{ {a - 1}} \bigg)\\ \\ </p><p></p><p> \red{ \boxed{ \bold{\therefore \: {cos} \: A = \bigg(\frac{a - 1}{ {a + 1}} \bigg)}}} \\ \\ {sin} \: A ={cos} \: A * {tan} \: A \\ \\ \hspace{25 pt}=\bigg(\frac{a - 1}{ {a + 1}} \bigg) * (2 √(a) )/(a - 1) \\ \\ \purple {\boxed { \bold{{sin} \: A = (2 √(a) )/(a + 1)}}}`