* 1 + tan^2 theta = sec^2 theta
sec^2 theta - tan^2 theta = 1
(sec theta + tan theta) (sec theta - tan theta) = 1 [since a^2 - b^2 = (a+b) (a-b) ]
{the product of two no.s is 1 that means they must be reciprocal of each other}
this implies that (sec theta + tan theta) and (sec theta - tan theta) are reciprocals of each other.
so let (sec theta + tan theta) = p ...............................eq. 1
and therefore (sec theta - tan theta) = 1/p...................eq.2
adding eq.1 and eq.2 we get,
2 sec theta = p +1/p
therefore , sec theta = p/2 + 1/ 2p
x + 1/4x = p/2 + 1/2p {since, sec theta = x + 1/4x (given)}
on compairing....
either p/2 = x or p/2 = 1/4x
if p/2 = x then,
p = 2x
this implies that sec theta + tan theta = 2x {since sec theta + tan theta = p}
or if p/2=1/4x then,
p = 1/2x
this implies that sec theta = tan theta = 1/2x