Question

If tan x =b/a
, then find the value of

root(a+b / a-b) + root(a-b / a+b)

Answer 1

tanx = b/a

$we \: know \\ tanx = \frac{b}{a} \\ b = a.tanx \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(i) \\ \\ \\ \: \: \: \: \: \sqrt{ \frac{a + b}{a - b} } + \sqrt{ \frac{a - b}{a + b} } \\ = \sqrt{ \frac{(a + b)(a + b)}{(a - b)(a + b)} } \\ + \sqrt{ \frac{(a - b)(a - b)}{(a - b)(a + b)} } \\ = \sqrt{ \frac{ {(a + b)}^{2} }{ {a}^{2} - {b}^{2} } } + \sqrt{ \frac{ {(a - b)}^{2} }{ {a}^{2} - {b}^{2} } } \\ = \frac{ \sqrt{ {(a + b)}^{2} } }{ \sqrt{ {a}^{2} - {b}^{2} } } + \frac{ \sqrt{ {(a - b)}^{2} } }{ {a}^{2} - {b}^{2} } \\ = \frac{a + b + a - b}{ \sqrt{ {a}^{2} - {b}^{2} } } = \frac{2a}{ \sqrt{ {a}^{2} - {b}^{2} } } \\ substituting \: (i) \: we \: get \\ = \frac{2a}{ \sqrt{ {a}^{2} - {(a.tanx)}^{2} } } \\ = \frac{2a}{ \sqrt{ {a}^{2} - {a}^{2}. {tan}^{2}x } } \\ = \frac{2a}{ \sqrt{ {a}^{2} (1 - {tan}^{2} x)} } \\ = \frac{2a}{ \sqrt{ {a}^{2} } \times \sqrt{(1 - {tan}^{2}x) } } \\ = \frac{2a}{a \times \sqrt{1 - {tan}^{2} x} } \\ = \frac{2}{ \sqrt{1 - {tan}^{2} x} }$