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# Given that $\tan 40\times \tan 6\theta =1$. Find the value of $\theta$, if $6\theta$ is given as a positive acute angle.

Last updated date: 02nd Aug 2024
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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 $\theta$.

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 $\theta$.