Question

If we have the expression \[\tan \theta = a \ne 0\], \[\tan 2\theta = b \ne 0\] and \[\tan \theta + \tan 2\theta = \tan 3\theta \] then
\[\left( 1 \right)\] \[a = b\]
\[\left( 2 \right)\] \[ab = 1\]
\[\left( 3 \right)\] \[\left( {a + b} \right) = 0\]
\[\left( 4 \right)\] \[b = 2a\]

Answer 1

Hint: We have to find the value of the given trigonometric expression \[\tan \theta + \tan 2\theta = \tan 3\theta \]. We solve this question using the concept of the various properties of the trigonometric functions. We should have the knowledge of the formula of the tangent of the sum of two angles. First we will simplify the right hand side of the given expression using the formula of the tangent of the sum of two angles and then we will simplify the expression and hence putting the values of the given tangent functions in the expression we will evaluate the required condition of the expression.

Complete step-by-step solution:
Given :
\[\tan \theta = a \ne 0\], \[\tan 2\theta = b \ne 0\], \[\tan \theta + \tan 2\theta = \tan 3\theta \]
We know that the formula of the tangent of the sum of two angles is given as :
\[\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a \times \tan b}}\]
We can write the right hand side as :
\[\tan 3\theta = \tan \left( {\theta + 2\theta } \right)\]
Now using the above formula, the right hand side of the expression can be written as :
\[\tan 3\theta = \dfrac{{\tan \theta + \tan 2\theta }}{{1 - \tan \theta \times \tan 2\theta }}\]
Substituting the values in the right hand side, we get the expression as :
\[\tan \theta + \tan 2\theta = \dfrac{{\tan \theta + \tan 2\theta }}{{1 - \tan \theta \times \tan 2\theta }}\]
Now on simplify the terms, we can write the expression as :
\[\left( {\tan \theta + \tan 2\theta } \right)\left( {1 - \tan \theta \times \tan 2\theta } \right) - \left( {\tan \theta + \tan 2\theta } \right) = 0\]
Taking \[\left( {\tan \theta + \tan 2\theta } \right)\] common, we get
\[\left( {\tan \theta + \tan 2\theta } \right)\left( {1 - \tan \theta \times \tan 2\theta - 1} \right) = 0\]
So, from the expression we can write the expression as :
\[\left( {\tan \theta + \tan 2\theta } \right) = 0\] or \[\left( {1 - \tan \theta \times \tan 2\theta - 1} \right) = 0\]
\[\left( {\tan \theta + \tan 2\theta } \right) = 0\] or \[\left( {\tan \theta \times \tan 2\theta } \right) = 0\]
Now, we will take the expression as :
\[\left( {\tan \theta + \tan 2\theta } \right) = 0\]
Putting the value in the expression, we get the value as :
\[\left( {a + b} \right) = 0\]
Thus, we conclude that the expression for the given trigonometric expression \[\tan \theta + \tan 2\theta = \tan 3\theta \] is \[\left( {a + b} \right) = 0\].
Hence, the correct option is \[\left( 3 \right)\].

Note: We will neglect the solution \[\left( {\tan \theta \times \tan 2\theta } \right) = 0\] as it is only possible when either \[\tan \theta = 0\] or \[\tan 2\theta = 0\] but it is given in the condition that \[\tan \theta \ne 0\] and \[\tan 2\theta \ne 0\]. So, as it contradicts the given conditions for the value we will neglect the solution.