Question

can anyone solve this question??if so..plss send answer as soon as possible....

Answer 1

heya..!!!?


given that:-

tanx = b/a


$\sqrt{ \frac{a + b}{a - b} } + \sqrt{ \frac{a - b}{a + b} } \\ \\ = > \frac{a + b + a - b}{ \sqrt{ {a}^{2} - {b}^{2} } } \\ \\ = > \frac{2a}{ \sqrt{ {a}^{2} - {b}^{2} } } \\ \\ = > \frac{ \frac{2a}{a} }{ \frac{ \sqrt{ {a}^{2} - {b}^{2} } }{a} } \\ \\ = > \frac{2}{ \sqrt{ \frac{ {a}^{2} - {b}^{2} }{ {a}^{2} } } } \\ \\ = > \frac{2}{ \sqrt{1 - {( \frac{ b }{a}) }^{2} } } \\ \\ = > \frac{2}{ \sqrt{1 - {tan}^{2} x} } \\ \\ = > \frac{2}{ \sqrt{ \frac{{cos}^{2}x - {sin}^{2}x}{ {cos}^{2}x } } } \\ \\ = > \frac{2 {cos}}{ \sqrt{cos2x} }$
. hence option (2) is correct