Question

Find the most general value of \[\theta \] satisfying the equation \[tan\theta + tan\left( {\dfrac{{3\pi }}{4} + \theta } \right) = 2\].
A. \[2n\pi \pm \dfrac{\pi }{3},n \in Z\]
B. \[n\pi \pm \dfrac{\pi }{3},n \in Z\]
C. \[2n\pi \pm \dfrac{\pi }{6},n \in Z\]
D. \[n\pi \pm \dfrac{\pi }{6},n \in Z\]

Answer 1

Hint:
In this question, we have to find the general value of \[\theta \]. We solve this using general values of trigonometric functions and trigonometric identities. We also know that the tan function is equal to the sin function divided by the cos function. By using trigonometric identities of cot functions and tan functions and general solution of trigonometric functions we can find the value of \[\theta \].

Formula used:
Write the basic trigonometric formula are given as
1. \[\tan \left( {\dfrac{\pi }{2} + \theta } \right) = - \cot \theta \]
2. \[\cot \left( {a - b} \right) = \dfrac{{\cot \left( a \right)\cot \left( b \right) - 1}}{{\cot \left( a \right) + \cot \left( b \right)}}\]
3. \[\tan \theta = \dfrac{1}{{\cot \theta }}\]

Complete step-by-step solution:
Given that \[tan\theta + tan\left( {\dfrac{{3\pi }}{4} + \theta } \right) = 2\]…(1)
Firstly, we will rewrite the equation (1) as
\[\tan \theta + \tan \left[ {\dfrac{\pi }{2} + \left( {\dfrac{\pi }{4} + \theta } \right)} \right] = 2\]
As we know, \[\tan \left( {\dfrac{\pi }{2} + \theta } \right) = - \cot \theta \].
Now, using this formula, we get
\[\tan \theta - \cot \left( {\dfrac{\pi }{4} - \theta } \right) = 2\]
Further, we will use the formula \[\cot \left( {a - b} \right) = \dfrac{{\cot \left( a \right)\cot \left( b \right) - 1}}{{\cot \left( a \right) + \cot \left( b \right)}}\], we get
\[\tan \theta - \left[ {\dfrac{{\cot \left( {\dfrac{\pi }{4}} \right)\cot \theta - 1}}{{\cot \left( {\dfrac{\pi }{4}} \right) + \cot \theta }}} \right] = 2\]
As we know that, \[\cot \dfrac{\pi }{4} = 1\].
So, we will simplify the above equation using the above trigonometric value, we get
\[\begin{array}{l}\tan \theta - \left[ {\dfrac{{1 \times \cot \theta - 1}}{{1 + \cot \theta }}} \right] = 2\\\tan \theta - \dfrac{{\left( {\cot \theta - 1} \right)}}{{\left( {1 + \cot \theta } \right)}} = 2\end{array}\]
Now, we will use this \[\tan \theta = \dfrac{1}{{\cot \theta }}\] formula in above equation, we get
\[\tan \theta - \dfrac{{\left( {1 - \tan \theta } \right)}}{{\left( {1 + \tan \theta } \right)}} = 2\]
Further, we will take the LCM and simplify the above equation by cancelling like terms, we get
\[\begin{array}{l}\tan \theta \left( {1 + \tan \theta } \right) - \left( {1 - \tan \theta } \right) = 2\left( {1 + \tan \theta } \right)\\\tan \theta + {\tan ^2}\theta - 1 + \tan \theta - 2 - 2\tan \theta = 0\\{\tan ^2}\theta = 3\end{array}\]
Further, we will simplify the above expression, we get
\[\tan \theta = \pm \sqrt 3 \]
As we know, that \[\tan \dfrac{\pi }{3} = \sqrt 3 \]
\[\tan \theta = \pm \tan \dfrac{\pi }{3}\]
Now we know that if then the general solution is
\[\therefore \theta = n\pi \pm \dfrac{\pi }{3},n \in Z\]
Therefore, if \[tan\theta + tan\left( {\dfrac{{3\pi }}{4} + \theta } \right) = 2\], then the general value of \[\theta \] is \[n\pi \pm \dfrac{\pi }{3}\].
Hence, option (B) is correct.

Note
Equations involving trigonometric functions of a variable are called trigonometric equations We can also solved it by using the formula \[\tan \left( {A + B} \right) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}\] The solutions of a trigonometric equation for \[0\leq x <2 \pi\] are called principle solutions. The expressions involving integers \[n\] which give all solutions of a trigonometric equation are called general solutions.