$\begin{align}  & y=\tan x \\ & y=\frac{\sin x}{\cos x} \\ & y+\delta y=\frac{\sin (x+\delta x)}{\cos (x+\delta x)} \\ & \delta y=\frac{\sin (x+\delta x)}{\cos (x+\delta x)}-\frac{\sin x}{\cos x} \\ & \delta y=\frac{\sin (x+\delta x)\cos x-\sin x\cos (x+\delta x)}{\cos x\cos (x+\delta x)} \\ & \delta y=\frac{\sin \left[ (x+\delta x)-x \right]}{\cos x\cos (x+\delta x)}=\frac{\sin \delta x}{\cos x\cos (x+\delta x)} \\ & \text{dividing through by }\delta x \\ & \frac{\delta y}{\delta x}=\frac{\sin \delta x}{\delta x\cdot \cos x\cos (x+\delta x)}=\frac{1}{\cos x\cos (x+\delta x)}\cdot \frac{\sin \delta x}{\delta x} \\ & \text{Proceeding to the limits }\delta x\to 0,\text{we have } \\ & \frac{dy}{dx}=\underset{\delta x\to 0}{\mathop{\lim }}\,\frac{1}{\cos x\cos (x+\delta x)}\cdot \underset{\delta x\to 0}{\mathop{\lim }}\,\frac{\sin \delta x}{x} \\ & \frac{dy}{dx}=\frac{1}{\cos x\cos x}\times 1=\frac{1}{{{\cos }^{2}}x}={{\sec }^{2}}x \\ & \text{Hence,}\frac{d}{dx}(\tan x)={{\sec }^{2}}x \\\end{align}$