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**Hint:**We start solving the problem by sending $\tan 40$ to the right side of the given equation $\tan 40\times \tan 6\theta =1$. We use the fact that $\cot x=\dfrac{1}{\tan x}$ to proceed through the problem. Now, we use the fact that $\cot x=\tan \left( \dfrac{\pi }{2}-x \right)$ to proceed further into the problem. Now, we use the fact that $\tan A=\tan B$, then A = B and make necessary calculations to get the required value of $\theta $.

**Complete step by step answer:**

Given that we have an equation $\tan 40\times \tan 6\theta =1$ and $6\theta $ is a positive acute angle. We need to find the value of $\theta $.

We have got the equation $\tan 40\times \tan 6\theta =1$.

$\tan 6\theta =\dfrac{1}{\tan 40}$ ---(1).

We know that $\cot x=\dfrac{1}{\tan x}$. We use this result in equation (1).

\[\tan 6\theta =\cot 40\] ---(2).

We know that $\cot x=\tan \left( \dfrac{\pi }{2}-x \right)$. We use this result in equation (2).

\[\tan 6\theta =\tan \left( \dfrac{\pi }{2}-40 \right)\] ---(3).

We know that if $\tan A=\tan B$, then A = B. We use this result in equation (3).

$6\theta =\dfrac{\pi }{2}-40$.

$\Rightarrow \theta =\dfrac{\pi }{2\times 6}-\dfrac{40}{6}$.

$\Rightarrow \theta =\dfrac{\pi }{12}-\dfrac{20}{3}$.

We have found the value of $\theta $ as $\dfrac{\pi }{12}-\dfrac{20}{3}$.

**∴ The value of $\theta $ is $\dfrac{\pi }{12}-\dfrac{20}{3}$.**

**Note:**We used $\cot x=\tan \left( \dfrac{\pi }{2}-x \right)$ in equation (2) as the given angle $6\theta $ is a positive acute angle. We know the fact that acute angle lies in between ${{0}^{o}}$ and ${{90}^{o}}$. If $6\theta $ is not a positive acute angle, then there will be more possibilities of answers for $\theta $. Similarly, we can expect the value of $\sin \theta $ or $\cos \theta $ after finding the value of $\theta $.

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