Trignometry: Find the value in terms of $ \ l \ $

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 1

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 $$x = 2\tan^{-1}\left(\frac{L - 1}{L + 1}\right), \quad L \ne -1.$$

Answer 2

$\displaystyle\sec x+\tan x=l\ \ \ \ (1)$ 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?

Answer 3

$$\sin x=\sqrt{1-\cos^2x}=\sqrt{1-\frac{1}{\sec^2x}}=\sqrt{\frac{\sec^2x-1}{\sec^2x}}\\\sec x \left(\sqrt{\frac{\sec^2x-1}{\sec^2x}}+1\right)=l\\\sec x=t\\t\sqrt{\frac{t^2-1}{t^2}}=l-t\\t^2\frac{t^2-1}{t^2}=l^2-2lt+t^2\\t^2-1-t^2=l^2-2lt\\2lt=l^2+1\\t=\frac{l^2+1}{2l}$$