Question

Solve for the general value of theta:

$\tan \theta \tan (120°-\theta )\times \tan (120°+\theta )=\frac{1}{\sqrt{3}}$

Answer 1

Step 1: Simplify the expression in L.H.S by using appropriate trigonometric identity

$\tan \theta \tan (120°-\theta )\times \tan (120°+\theta )=\frac{1}{\sqrt{3}}$

Let us express $\tan 120°$ as $\tan (180-60)°$

On substituting the value of $\tan 120°$ from above in the given equation

$\tan \theta \times \tan \left(\left({180}^{\circ }-{60}^{\circ }\right)-\theta \right)\times \tan \left(\left({180}^{\circ }-{60}^{\circ }\right)+\theta \right)=\frac{1}{\sqrt{3}}$

On rearranging the terms we get

$\tan \theta \times \tan \left({180}^{\circ }+\left(-\theta -{60}^{\circ }\right)\right)\times \tan \left({180}^{\circ }+\left(\theta -{60}^{\circ }\right)\right)=\frac{1}{\sqrt{3}}$

From trigonometric identity, we know that

$\tan \left({180}^{\circ }+\theta \right)=\tan \theta$

So the above equation becomes

$\tan \theta \times \tan \left(-\theta -{60}^{\circ }\right)\times \tan \left(\theta -{60}^{\circ }\right)=\frac{1}{\sqrt{3}}$

$-\tan \theta \times \tan \left(\theta +{60}^{\circ }\right)\times \tan \left(\theta -{60}^{\circ }\right)=\frac{1}{\sqrt{3}}$

On expanding using the trigonometric identity, we get

Applying these in our equation $-\tan \theta \times \tan \left(\theta +{60}^{\circ }\right)\times \tan \left(\theta -{60}^{\circ }\right)=\frac{1}{\sqrt{3}}$, we get

Step 2: Use compound angle formula of tangent

$-\tan \theta \times \left[\frac{\tan \theta +\tan 60}{1-\tan \theta \tan 60}\right]\times \left[\frac{\tan \theta -\tan 60}{1+\tan \theta \tan 60}\right]=\frac{1}{\sqrt{3}}$ $\left[\because \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A.\tan B},\tan \left(A-B\right)=\frac{\tan A-\tan B}{1+\tan A.\tan B}\right]$

We know that $\tan 60°=\sqrt{3}$

$-\tan \theta \times \left[\frac{\tan \theta +\sqrt{3}}{1-\tan \theta \times \sqrt{3}}\right]\times \left[\frac{\tan \theta -\sqrt{3}}{1+\tan \theta \times \sqrt{3}}\right]=\frac{1}{\sqrt{3}}$

$$\begin{gathered} -\tan \theta \times \frac{\left({\tan }^2\theta -3\right)}{\left(1-3{\tan }^2\theta \right)}=\frac{1}{\sqrt{3}} \\ \Rightarrow \frac{\left(3\tan \theta -{\tan }^3\theta \right)}{\left(1-3{\tan }^2\theta \right)}=\frac{1}{\sqrt{3}} \end{gathered}$$

We have learnt that $\tan 3\theta =\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }$

$$\begin{gathered} \tan 3\theta =\frac{1}{\sqrt{3}} \\ \Rightarrow \tan 3\theta =\tan 30°\left[\because \tan 30°=\frac{1}{\sqrt{3}}\right] \\ \Rightarrow \tan 3\theta =\tan \frac{\mathrm{\pi }}{6} \end{gathered}$$

Step 3: Find the value of $\mathrm{\theta }$

As we know that if $\tan x=\tan \theta$, then the general solution is $x=n\mathrm{\pi }+\mathrm{\theta }$

$3\theta =n\mathrm{\pi }+\frac{\mathrm{\pi }}{6},n\in Z$

$\theta =\frac{\mathrm{n\pi }}{3}+\frac{\mathrm{\pi }}{18},n\in Z$

Hence, $\theta =\frac{\mathrm{n\pi }}{3}+\frac{\mathrm{\pi }}{18},n\in Z$ is the solution of given trigonometric equation.