Hello Rahul,
Let's begin with:
$$\tanh\left(\frac{x}{2} \right)=\frac{\sinh\left(\frac{x}{2} \right)}{\cosh\left(\frac{x}{2} \right)}$$
Next, we can employ the definitions:
$$\sinh(u)\equiv\frac{e^u-e^{-u}}{2}$$
$$\cosh(u)\equiv\frac{e^u+e^{-u}}{2}$$
To obtain:
$$\tanh\left(\frac{x}{2} \right)=\frac{\dfrac{e^{\frac{x}{2}}-e^{-\frac{x}{2}}}{2}}{\dfrac{e^{\frac{x}{2}}+e^{-\frac{x}{2}}}{2}}$$
Multiplying the right side by $$1=\frac{2}{2}$$ we obtain:
$$\tanh\left(\frac{x}{2} \right)=\frac{e^{\frac{x}{2}}-e^{-\frac{x}{2}}}{e^{\frac{x}{2}}+e^{-\frac{x}{2}}}$$
Squaring both sides, we obtain:
$$\tanh^2\left(\frac{x}{2} \right)=\frac{e^{x}+e^{-x}-2}{e^{x}+e^{-x}+2}$$
Dividing each term in the numerator and denominator on the right by $2$ we may write:
$$\tanh^2\left(\frac{x}{2} \right)=\frac{\dfrac{e^{x}+e^{-x}}{2}-1}{\dfrac{e^{x}+e^{-x}}{2}+1}$$
Using the definition for the hyperbolic cosine function, we obtain:
$$\tanh^2\left(\frac{x}{2} \right)=\frac{\cosh(x)-1}{\cosh(x)+1}$$
Now, we are given $$\cosh(x)=\sec(\theta)$$, and so we may write:
$$\tanh^2\left(\frac{x}{2} \right)=\frac{\sec(\theta)-1}{\sec(\theta)+1}$$
Multiply the right side by $$1=\frac{\cos(\theta)}{\cos(\theta)}$$ to get:
$$\tanh^2\left(\frac{x}{2} \right)=\frac{1-\cos(\theta)}{1+\cos(\theta)}$$
In the numerator use the double-angle identity for cosine $$\cos(2u)=1-2\sin^2(u)$$ and in the denominator use the double-angle identity for cosine $$\cos(2u)=2\cos^2(u)-1$$ to get:
$$\tanh^2\left(\frac{x}{2} \right)=\frac{1-\left(1-2\sin^2\left(\dfrac{\theta}{2} \right) \right)}{1+\left(2\cos^2\left(\dfrac{\theta}{2} \right)-1 \right)}=\frac{2\sin^2\left(\dfrac{\theta}{2} \right)}{2\cos^2\left(\dfrac{\theta}{2} \right)}$$
Hence, we have:
$$\tanh^2\left(\frac{x}{2} \right)=\tan^2\left(\dfrac{\theta}{2} \right)$$
And this is what we were asked to show.