If sec = x + 14x ( 0^ < < 90^ ) then sec + tan is equal to?

Answer: Option B

$$\begin{array}{rl}& sec\theta =x+\frac{1}{4x}\\ & \text{tan}\theta =\sqrt{\text{sec}\theta -1}\\ & \text{tan}\theta =\sqrt{{\left[\frac{4x^2+1}{4x}\right]}^2-1}\\ & \text{tan}\theta =\sqrt{\frac{{\left(4x^2+1\right)}^2-{\left(4x\right)}^2}{{\left(4x\right)}^2}}\\ & \text{tan}\theta =\sqrt{\frac{16x^4+1+8x^2-16x^2}{{\left(4x\right)}^2}}\\ & \text{tan}\theta =\sqrt{\frac{16x^4+1-8x^2}{{\left(4x\right)}^2}}\\ & \text{tan}\theta =\sqrt{\frac{{\left(4x^2-1\right)}^2}{{\left(4x\right)}^2}}\\ & \text{tan}\theta =\frac{\left(4x^2-1\right)}{4x}\\ & \therefore sec\theta +\text{tan}\theta \\ & =\frac{4x^2+1}{4x}+\frac{4x^2-1}{4x}\\ & =\frac{4x^2+1+4x^2-1}{4x}\\ & =\frac{8x^2}{4x}\\ & =2x\\ & \\ & \mathbf{Alternate}:\\ & sec\theta =x+\frac{1}{4x}\\ & \text{Put }x=1\\ & sec\theta =1+\frac{1}{4}=\frac{5}{4}=\frac{\text{H}}{\text{B}}\\ & \tan \theta =\frac{\text{P}}{\text{B}}=\frac{3}{4}\\ & \text{Now, }\\ & sec\theta +\text{tan}\theta \\ & =\frac{5}{4}+\frac{3}{4}\\ & =\frac{5+3}{4}\\ & =\frac{8}{4}\\ & =2\times 1\\ & =2x\left(x=1\right)\end{array}$$