Equation involving Tan and Sec

Sec^2(x) - Tan^2(x) = 1 , this means that

(Tan(x) + Sec(x))(Sec(x) - Tan(x)) = 1 ------------------(1)

As it is given that Tan(x) + Sec(x) = 22/7, it implies that due to equation (1) above

Sec(x) - Tan(x) = 7/22 -------------------(2)

Solving for Tan(x) by subtracting (2) from (1) yields

2* Tan(x) = 22/7 - 7/22 = 435/154 or Tan(x) = 435/308 = 1.412

Taking $\tan x=y, y+\sqrt {1+y^2}=\dfrac{22}{7}\implies y=\dfrac{(\dfrac{22}{7})^2-1}{2\times \dfrac{22}{7}}\approx \boxed {1.41164}$

We can use half-angle tangent substitution to solve this problem as follows.

$\begin{aligned} \tan x + \sec x & = \frac {22}7 & \small \blue{\text{Let }t=\tan \frac x2} \\ \frac {2t}{1-t^2} + \frac {1+t^2}{1-t^2} & = \frac {22}7 \\ \frac {(1+t)^2}{(1-t)(1+t)} & = \frac {22}7 \\ \frac {1+t}{1-t} & = \frac {22}7 \\ 7 + 7t & = 22-22t \\ \implies t & = \frac {15}{29} \\ \implies \tan x & = \frac {2t}{1-t^2} = \frac {29(2)(15)}{29^2 - 15^2} \approx \boxed{1.412} \end{aligned}$