Question

If $\sec x-\tan x=\frac{2}{3},$ then tan x = ___________.

Answer 1

$$\begin{gathered} \mathrm{Given}\sec x-\tan x=\frac{2}{3}\left(1\right) \\ \mathrm{Since}{\sec }^{2}x-{\tan }^{2}x=1 \\ \Rightarrow \left(\sec x-\tan x\right)\left(\sec x+\tan x\right)=1 \\ \Rightarrow \frac{2}{3}\left(\sec x+\tan x\right)=1 \\ \Rightarrow \sec x+\tan x=\frac{3}{2}\left(2\right) \end{gathered}$$
Subtracting (1) from (2)
$$\begin{gathered} \sec x+\tan x-\sec x+\tan x=\frac{3}{2}-\frac{2}{3} \\ \Rightarrow 2\tan x=\frac{9-4}{6} \\ \Rightarrow 2\tan x=\frac{5}{6} \\ \Rightarrow \tan x=\frac{5}{12} \end{gathered}$$