Find the value of tan x, if sec x + tanx = $\sqrt{3}$ and x lies between 0^o and 90^o.

$1/\sqrt{3}$

We are given that ,

sec x + tan x = \(\sqrt {3 }\)   ----(1)

{ using , sec² - tan²x = 1   so,  secx - tanx = \(\frac{1}{sec x + tan x}\)  }

secx - tanx = \(\frac{1}{ √3 }\)     ----(2)

On adding equation 1 and 2.

2 secx = \(\sqrt {3 }\)  + \(\frac{1}{ √3 }\)

2 secx =  \(\frac{4}{ √3 }\)

secx =  \(\frac{2}{ √3 }\)

{ we know, sec 30º =  \(\frac{2}{ √3 }\) }

So, x = 30º 

Now,

tanx

= tan 30º 

= \(\frac{1}{ √3 }\)