Question

If we have an equation as $\tan \theta + \tan \left( {\theta + \dfrac{\pi }{3}} \right) + \tan \left( {\theta + \dfrac{{2\pi }}{3}} \right) = 3$ , then which of the following is equal to $1$ ?
$\left( 1 \right)$ $\tan 2\theta$
$\left( 2 \right)$ $\tan 3\theta$
$\left( 3 \right)$ ${\tan ^2}\theta$
$\left( 4 \right)$ ${\tan ^3}\theta$

Answer 1

We have to evaluate the expression such that we get the relation of the trigonometric expression which is equal to $1$ . We solve the question by using trigonometric identities and the expression of the triple angle of a trigonometric function . We use the formula of tan of sum of two angles and after expanding the formula and putting the values and simplifying we get the required relation of trigonometric expression which is equal to $1$ .

Complete step-by-step solution:
Given : $\tan \theta + \tan \left( {\theta + \dfrac{\pi }{3}} \right) + \tan \left( {\theta + \dfrac{{2\pi }}{3}} \right) = 3$
As we know that the formulas of sum and difference of two angles of a tangent function is given as :$\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a \times \tan b}}$ and $\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a \times \tan b}}$
We expand the terms in the above expression .
After expanding the terms , we get
$\tan \theta + \dfrac{{\tan \theta + \tan \dfrac{\pi }{3}}}{{1 - \tan \theta \times \tan \dfrac{\pi }{3}}} + \dfrac{{\tan \theta + \tan \dfrac{{2\pi }}{3}}}{{1 - \tan \theta \times \tan \dfrac{{2\pi }}{3}}} = 3$
As , we know that the value of tangent function is given as :
$\tan \dfrac{\pi }{3} = \sqrt 3$ and $\tan \dfrac{{2\pi }}{3} = - \sqrt 3$
Now , putting the values in the expression , we get the expression as :
$\tan \theta + \dfrac{{\tan \theta + \sqrt 3 }}{{1 - \sqrt 3 \tan \theta }} + \dfrac{{\tan \theta - \sqrt 3 }}{{1 + \sqrt 3 \tan \theta }} = 3$
On , taking L.C.M. and simplifying it further , we get the expression as :
$\dfrac{{\tan \theta \left( {1 - 3{{\tan }^2}\theta } \right) + \left( {\tan \theta + \sqrt 3 } \right)\left( {1 + \sqrt 3 \tan \theta } \right) + \left( {\tan \theta - \sqrt 3 } \right)\left( {1 - \sqrt 3 \tan \theta } \right)}}{{1 - 3{{\tan }^2}\theta }} = 3$
Further simplifying , we get
$\dfrac{{\left( {\tan \theta - 3{{\tan }^3}\theta } \right) + \left( {\tan \theta + \sqrt 3 } \right) + \left( {3\tan \theta + \sqrt 3 {{\tan }^2}\theta } \right) + \left( {\tan \theta - \sqrt 3 } \right) + \left( {3\tan \theta - \sqrt 3 {{\tan }^2}\theta } \right)}}{{1 - 3{{\tan }^2}\theta }} = 3$
After cancelling the terms , we get the expression as :
$\dfrac{{9\tan \theta - 3{{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }} = 3$
$\dfrac{{3\left( {3\tan \theta - {{\tan }^3}\theta } \right)}}{{1 - 3{{\tan }^2}\theta }} = 3$
cancelling the terms , we can write the expression as :
$\dfrac{{3\tan \theta - {{\tan }^3}\theta }}{{1 - 3{{\tan }^2}\theta }} = 1$
We also know that , the formula of triple angle of tangent function is given as :
$\tan 3x = \dfrac{{3\tan x - {{\tan }^3}x}}{{1 - 3{{\tan }^2}x}}$
Using this formula , we get the expression as :
$\tan 3\theta = 1$
Thus , we conclude that from the given expression $\tan 3\theta = 1$ .
Hence the correct option is $\left( 2 \right)$ .

Note: All the trigonometric functions are classified into two categories or types as either sine function or cosine function . All the functions which lie in the category of sine functions are sin , cosec and tan functions on the other hand the functions which lie in the category of cosine functions are cos , sec and cot functions . The trigonometric functions are classified into these two categories on the basis of their property which is stated as : when the value of angle is substituted by the negative value of the angle then we get the negative value for the functions in the sine family and a positive value for the functions in the cosine family .
We have various trigonometric formulas used to solve the problem
The various trigonometric formulas used :
$\sin 2x = 2\sin x\cos x$
$\cos 2x = {\cos ^2}x - {\sin ^2}x$
$\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}$
$\sin 3x = 3\sin x - 4{\sin ^3}x$
$\cos 3x = 4{\cos ^3}x - 3\cos x$