Question

If \[{\tan ^2}\theta - \left( {1 + \sqrt 3 } \right)\tan \theta + \sqrt 3 = 0\], then what is the value of \[\theta \]?
A.\[n\pi + \dfrac{\pi }{4},n\pi + \dfrac{\pi }{3}\]
B.\[n\pi - \dfrac{\pi }{4},n\pi + \dfrac{\pi }{3}\]
C.\[n\pi + \dfrac{\pi }{4},n\pi - \dfrac{\pi }{3}\]
D.\[n\pi - \dfrac{\pi }{4},n\pi - \dfrac{\pi }{3}\]

Answer 1

Hint: In solving the given equation first we will apply the distributive property to remove the parenthesis. Then apply the factorization method and equate each factor with zero. From this, we will calculate the value of \[\theta\].

Formula used: We will use the general solution of \[\tan \theta \] which is given by
If \[\tan \theta = \tan \alpha \], then the general solution will be, \[\theta = n\pi + \alpha \].

Complete step-by-step solution:
Given \[{\tan ^2}\theta - \left( {1 + \sqrt 3 } \right)\tan \theta + \sqrt 3 = 0\]
Now we will open the parenthesis and we will simplify, we will get,
\[ \Rightarrow {\tan ^2}\theta - \tan \theta - \sqrt 3 \tan \theta + \sqrt 3 = 0\]
Now we will take out the common like terms, we will get,
\[ \Rightarrow \tan \theta \left( {\tan \theta - 1} \right) - \sqrt 3 \left( {\tan \theta - 1} \right) = 0\]
Now we will again take out the common like terms, we will get,
\[ \Rightarrow \left( {\tan \theta - \sqrt 3 } \right)\left( {\tan \theta - 1} \right) = 0\]
Now we will equate each term to 0, first we will take first term, then we will get,
\[ \Rightarrow \tan \theta - \sqrt 3 = 0\]
Now add\[\sqrt 3 \] to both sides, then we will get,
\[ \Rightarrow \tan \theta - \sqrt 3 + \sqrt 3 = 0 + \sqrt 3 \]
Now we will simplify, then we will get,
\[ \Rightarrow \tan \theta = \sqrt 3 \]
Now we will use the trigonometric tables, we will get,
\[ \Rightarrow \tan \theta = \tan \dfrac{\pi }{3}\]
Now will take second term, we will get,
\[ \Rightarrow \tan \theta - 1 = 0\]
Now add 1 to both sides, then we will get,
\[ \Rightarrow \tan \theta - 1 + 1 = 0 + 1\]
Now we will simplify, then we will get,
\[ \Rightarrow \tan \theta = 1\]
Now we will use the trigonometric tables, we will get,
\[ \Rightarrow \tan \theta = \tan \dfrac{\pi }{4}\]
Now using the general solution for \[\tan \theta \] which is given by
If \[\tan \theta = \tan \alpha \] then general solution will be \[\theta = n\pi + \alpha \], we will get,
\[ \Rightarrow \theta = n\pi + \dfrac{\pi }{4},n\pi + \dfrac{\pi }{3}\]

The correct option is A.

Note: Students often confused with the general solution of \[\tan \theta = \tan \alpha \]. Sometimes they applied the formula \[\theta = n\pi \pm \alpha \] which is general solution of \[\tan 2\theta = \tan 2\alpha \].