Question: If $ \ \tan x + \sec x \ = \ l \ , $
find the value of secant x in terms of $ \ l \ $ .
I do not know how to solve it but i do know that the basic trig identities are involved.
Pls help me!!!

Answer: 
$\displaystyle\sec x+\tan x=l\ \ \ \ (1)$
As $\displaystyle \sec ^2x-\tan^2x=1, (\sec x-\tan x)(\sec x+\tan x)=1$

$\displaystyle \sec x-\tan x=\frac1{(\sec x+\tan x)}=\frac1l$

Can you find $\sec x$ from here?
$$\sin x=\sqrt{1-\cos^2x}=\sqrt{1-\frac{1}{\sec^2x}}=\sqrt{\frac{\sec^2x-1}{\sec^2x}}\\\sec x \left(\sqrt{\frac{\sec^2x-1}{\sec^2x}}+1\right)=l\\\sec x=t\\t\sqrt{\frac{t^2-1}{t^2}}=l-t\\t^2\frac{t^2-1}{t^2}=l^2-2lt+t^2\\t^2-1-t^2=l^2-2lt\\2lt=l^2+1\\t=\frac{l^2+1}{2l}$$