Question

Solve $tan\text{ }\theta \text{ }+tan\left( \theta +\pi /3 \right)+\text{ }tan\left( \theta +2\pi /3 \right)=3$.

Answer 1

The basic identities which are used in solving trigonometric equations are called " The Fundamental Trigonometric identities”. Some of these identities are reciprocal identities, Pythagorean and the quotient identities. For solving equations, we generally use combinations of these identities. In order to reduce a complex form of expression to its simplest form, these identities play an important role. The best way to use these identities in solutions is memorization of these identities. Also, we can memorise the value of trigonometric expressions for some common angles such as, etc. We can also use these Fundamental identities to prove other identities in trigonometry. In this question, we will solve the equation by using the formula for identity tan(A+B) and tan(A-B).

Complete step by step solution:
The given equation is
$tan\text{ }\theta \text{ }+tan\left( \theta +\pi /3 \right)+\text{ }tan\left( \theta +2\pi /3 \right)=3$
We know the formula for $tan\text{ }\left( A+B \right)$
$tan\left( A+B \right)=\dfrac{tanA+tanB}{1-tanAtanB}$
Now we will use this formula to simplify $tan\left( \theta +\pi /3 \right)$ and $tan\left( \theta +2\pi /3 \right)$
$tan\left( \theta +\pi /3 \right)=\dfrac{tan\theta +tan\pi /3}{1-tan\theta tan\pi /3}$
$tan\left( \theta +2\pi /3 \right)=\dfrac{tan\theta +tan2\pi /3}{1-tan\theta tan2\pi /3}$
Using these in the above equation , we get
$\tan \theta +\dfrac{tan\theta +tan\pi /3}{1-tan\theta tan\pi /3}+\dfrac{tan\theta +tan2\pi /3}{1-tan\theta tan2\pi /3}=3$
We already know the value of $tan\pi /3$ which is $\surd 3$ and $\tan \dfrac{2\pi }{3}=-\sqrt{3}$
Now using these values , we get
$\begin{aligned} & \tan \theta +\dfrac{tan\theta +\sqrt{3}}{1-\sqrt{3}tan\theta }+\dfrac{tan\theta -\sqrt{3}}{1+\sqrt{3}tan\theta }=3 \\\ & \Rightarrow \tan \theta +\dfrac{8tan\theta }{1-3ta{{n}^{2}}\theta }=3 \\\ & \Rightarrow \dfrac{\tan \theta (1-3{{\tan }^{2}}\theta )+8tan\theta }{1-3ta{{n}^{2}}\theta }=3 \\\ & \Rightarrow \dfrac{3(3\tan \theta -{{\tan }^{3}}\theta )}{1-3ta{{n}^{2}}\theta }=3 \\\ & \Rightarrow 3\tan 3\theta =3 \\\ & \Rightarrow \tan 3\theta =1 \\\ & \Rightarrow 3\theta =n\pi +\dfrac{\pi }{4} \\\ & \Rightarrow \theta =\dfrac{n\pi }{3}+\dfrac{\pi }{12} \\\ \end{aligned}$

Hence we get the general solution of the equation which is $\dfrac{n\pi }{3}+\dfrac{\pi }{12}$

Note:
In this question , we are not asked to give the solution in the range or any particular interval, so we will consider all the points for which the equation holds. This solution is called a general solution.