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inysia [295]
3 years ago
5

(If tan A=)PLZ ANSWER IT FAST...​

Mathematics
1 answer:
svetoff [14.1K]3 years ago
5 0

Answer:

See explanation

Step-by-step explanation:

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)\\  \\

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)\\  \\  \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}}}}

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