1. Given integral

We need to evaluate

$$\int_0^{\pi/3} \frac{\tan\theta}{\sqrt{2k\,\sec\theta}}\,d\theta =1-\frac1{\sqrt2}, \qquad k>0.$$

We simplify the integrand:

$$\frac{\tan\theta}{\sqrt{2k\sec\theta}} =\frac{1}{\sqrt{2k}}\cdot \frac{\tan\theta}{\sqrt{\sec\theta}}.$$

Now,

$$\frac{\tan\theta}{\sqrt{\sec\theta}} =\frac{\sin\theta/\cos\theta}{1/\sqrt{\cos\theta}} =\frac{\sin\theta}{\sqrt{\cos\theta}}.$$

So the integral becomes

$$\frac{1}{\sqrt{2k}}\int_0^{\pi/3} \frac{\sin\theta}{\sqrt{\cos\theta}}\,d\theta.$$

2. Substitution

Let

$$u=\cos\theta \quad \Rightarrow \quad d\nu=-\sin\theta\,d\theta.$$

Then

$$\int \frac{\sin\theta}{\sqrt{\cos\theta}}\,d\theta = -\int \nu^{-1/2}\,d\nu = -2\sqrt{\nu} = -2\sqrt{\cos\theta}.$$

Hence,

$$\int_0^{\pi/3} \frac{\sin\theta}{\sqrt{\cos\theta}}\,d\theta =\left[-2\sqrt{\cos\theta}\right]_0^{\pi/3}.$$

Now,

$$\cos\frac{\pi}{3}=\frac12, \qquad \cos 0=1.$$

So,

$$\left[-2\sqrt{\cos\theta}\right]_0^{\pi/3} =-2\sqrt{\frac12}+2 =2-\frac{2}{\sqrt2} =2-\sqrt2.$$

Thus the given integral equals

$$\frac{1}{\sqrt{2k}}(2-\sqrt2).$$

3. Use the given value

We are given

$$\frac{2-\sqrt2}{\sqrt{2k}}=1-\frac1{\sqrt2}.$$

Notice that

$$2-\sqrt2=\sqrt2\left(\sqrt2-1\right),$$

and

$$1-\frac1{\sqrt2}=\frac{\sqrt2-1}{\sqrt2}.$$

So,

$$\frac{\sqrt2(\sqrt2-1)}{\sqrt{2k}}=\frac{\sqrt2-1}{\sqrt2}.$$

Since $\sqrt2-1\neq 0$, cancel it:

$$\frac{\sqrt2}{\sqrt{2k}}=\frac{1}{\sqrt2}.$$

Cross-multiplying,

$$\sqrt2\cdot \sqrt2=\sqrt{2k} \Rightarrow 2=\sqrt{2k}.$$

Squaring both sides,

$$4=2k \Rightarrow k=2.$$

4. Check options

• A: $4$ ❌
• B: $\frac12$ ❌
• C: $1$ ❌
• D: $2$ ✅

Therefore, the correct answer is

$$\boxed{2}.$$