<s>[INST] Trignometry: Find the value in terms of $ \ l \ $ 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!!!
 [/INST] Simply use the definitions:
$$\tan x + \sec x = \frac{\sin x}{\cos x} + \frac{1}{\cos x} = \frac{1 + \sin x}{\cos x}.$$
So we seek $x$ such that
$$1 + \sin x = L \cos x \quad \text{or} \quad L \cos x - \sin x = 1.$$
Let $t = \tan(x/2)$. Then
$$\sin x = 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right) = 2\frac{\sin(x/2)}{\cos(x/2)}\cos^2\left(\frac{x}{2}\right) = 2\frac{\tan(x/2)}{\sec^2(x/2)} = \color{red}{\frac{2t}{1 + t^2}}.$$
Similarly,
$$\cos x = \cos^2\left(\frac{x}{2}\right) - \sin^2\left(\frac{x}{2}\right) = \cos^2\left(\frac{x}{2}\right)\left(1 - \frac{\sin^2(x/2)}{\cos^2(x/2)}\right) = \frac{1 - \tan^2(x/2)}{\sec^2(x/2)} = \color{red}{\frac{1 - t^2}{1 + t^2}}.$$
To obtain the expressions in $\color{red}{\text{red}}$ just draw a right triangle with angle $= x/2$.
Therefore, $L \cos x - \sin x = 1$ transforms into
$$L\frac{1 - t^2}{1 + t^2} - \frac{2t}{1 + t^2} - 1 = 0.$$
If you solve for $t$ you should get $t = -1$ or $t = (L - 1)/(L + 1)$. So
$$x = -\frac{\pi}{2} \quad \text{or} \quad x = 2\tan^{-1}\left(\frac{L - 1}{L + 1}\right).$$
Clearly, $x \ne -\pi/2$ because $\cos(-\pi/2) = 0$, so
$$x = 2\tan^{-1}\left(\frac{L - 1}{L + 1}\right), \quad L \ne -1.$$
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