If tanθ + secθ = x - 2x + 2, then what is the value of cosθ?

Answer: Option C

$$\begin{array}{rl}& \tan \theta +\sec \theta =\frac{x-2}{x+2}........\left(\text{i}\right)\\ & \frac{\left(\sec \theta +\tan \theta \right)}{1}\times \frac{\left(\sec \theta -\tan \theta \right)}{\left(\sec \theta -\tan \theta \right)}=\frac{x-2}{x+2}\\ & \frac{1}{\sec \theta -\tan \theta }=\frac{x-2}{x+2}\\ & \left(\because {\sec }^2\theta -{\tan }^2\theta =1\right)\\ & \sec \theta -\tan \theta =\frac{x+2}{x-2}........\left(\text{ii}\right)\\ & \text{By equation }\left(\text{ii}\right)+\left(\text{i}\right)\text{ we get}\\ & 2\sec \theta =\frac{x+2}{x-2}+\frac{x-2}{x+2}\\ & 2\sec \theta =\frac{{\left(x+2\right)}^2+{\left(x-2\right)}^2}{\left(x^2-2^2\right)}\\ & 2\sec \theta =\frac{\left[x^2+4+2x+x^2+4-2x\right]}{\left(x^2-4\right)}\\ & 2\sec \theta =\frac{2\left(x^2+4\right)}{\left(x^2-4\right)}\\ & \sec \theta =\frac{x^2+4}{x^2-4}\\ & \frac{1}{\cos \theta }=\frac{x^2+4}{x^2-4}\\ & \cos \theta =\frac{x^2-4}{x^2+4}\end{array}$$