If 1 + + 1 + = 1 3 , 0° < θ < 90° then the value of (tanθ + secθ)-1 is?

Answer: Option C

$$\begin{array}{rl}& \frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }=\frac{1}{\sqrt{3}}\\ & \frac{{\sin }^2\theta +{\left(1+\cos \theta \right)}^2}{\sin \theta \left(1+\cos \theta \right)}=\frac{1}{\sqrt{3}}\\ & \frac{{\sin }^2\theta +1+{\cos }^2\theta +2\cos \theta }{\sin \theta \left(1+\cos \theta \right)}=\frac{1}{\sqrt{3}}\\ & \frac{{\sin }^2\theta +{\cos }^2\theta +1+2\cos \theta }{\sin \theta \left(1+\cos \theta \right)}=\frac{1}{\sqrt{3}}\\ & \frac{2\left(1+\cos \theta \right)}{\sin \theta \left(1+\cos \theta \right)}=\frac{2}{\sqrt{3}}\\ & \sin \theta =\frac{\sqrt{3}}{2}\\ & \theta =60^{\circ }\\ & \sec \theta -\tan \theta \\ & =\sec 60^{\circ }-\tan 60^{\circ }\\ & =2-\sqrt{3}\end{array}$$