1. Given relation

We have $f(\cos 4\theta)=\frac{2}{2-\sec^2\theta}, \qquad \theta\in\left(0,\frac\pi4\right)\cup\left(\frac\pi4,\frac\pi2\right).$

We need to find the possible value(s) of $f\left(\frac13\right).$

So we must solve $\cos 4\theta=\frac13$ and then evaluate $\frac{2}{2-\sec^2\theta}.$

2. Express the RHS in terms of $\cos 2\theta$

Since $\sec^2\theta=\frac{1}{\cos^2\theta},$ we get

Multiply numerator and denominator by $\cos^2\theta$:

Now use $2\cos^2\theta-1=\cos 2\theta,$ so

Hence

Thus, $f(\cos 4\theta)=1+\sec 2\theta.$

3. Use $\cos 4\theta=\frac13$

Let $x=\cos 2\theta.$ Then $\cos 4\theta=2\cos^2 2\theta-1=2x^2-1.$

Given $2x^2-1=\frac13,$ so $2x^2=\frac43$ $x^2=\frac23.$ Therefore $x=\pm\sqrt{\frac23}.$

That is, $\cos 2\theta=\pm\sqrt{\frac23}.$

Both signs are possible because:

• if $\theta\in\left(0,\frac\pi4\right)$, then $2\theta\in\left(0,\frac\pi2\right)$, so $\cos 2\theta>0$;
• if $\theta\in\left(\frac\pi4,\frac\pi2\right)$, then $2\theta\in\left(\frac\pi2,\pi\right)$, so $\cos 2\theta<0$.

So both values occur.

4. Compute $f\left(\frac13\right)$

We have

Since $\cos 2\theta=\pm\sqrt{\frac23},$ we get

Hence

So the possible values are $1-\sqrt{\frac32} \quad \text{and} \quad 1+\sqrt{\frac32}.$

5. Check options

• A: $1-\sqrt{\frac32}$ ✅
• B: $1+\sqrt{\frac32}$ ✅
• C: $1-\sqrt{\frac23}$ ❌
• D: $1+\sqrt{\frac23}$ ❌

6. Comparison with stored answer

Stored correct answer: A, B.

Our derived answer is also A, B. So they agree.