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Question: If a trigonometric equation is given as \(\tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\c...

If a trigonometric equation is given as tan⁡θ⋅tan⁡(1200−θ)⋅tan⁡(1200+θ)=13\tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\dfrac{1}{\sqrt{3}} then θ=\theta = .
(a) nπ3−π12,n∈z\dfrac{n\pi }{3}-\dfrac{\pi }{12},n\in z
(b) nπ3−π18,n∈z\dfrac{n\pi }{3}-\dfrac{\pi }{18},n\in z
(c) nπ3+π18,n∈z\dfrac{n\pi }{3}+\dfrac{\pi }{18},n\in z
(d) nπ3+π12,n∈z\dfrac{n\pi }{3}+\dfrac{\pi }{12},n\in z

Explanation

Solution

Hint: For solving this question we will use the formulas like tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B} and tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B} for proving tan⁡θ⋅tan⁡(1200−θ)⋅tan⁡(1200+θ)=tan⁡3θ\tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\tan 3\theta . After that, we will use the result of the solution of equation tan⁡x=tan⁡y\tan x=\tan y to write the final answer.

Complete step-by-step solution -
Given:
It is given that, tan⁡θ⋅tan⁡(1200−θ)⋅tan⁡(1200+θ)=13\tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\dfrac{1}{\sqrt{3}} and we have to find the suitable values of θ\theta .
Now, before we proceed we should know the following formulas:
tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B..................(1) tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B..................(2) tan⁡1200=−3......................................(3) (a+b)(a−b)=a2−b2..........................(4) tan⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ........................(5) \begin{aligned} & \tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}..................\left( 1 \right) \\\ & \tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}..................\left( 2 \right) \\\ & \tan {{120}^{0}}=-\sqrt{3}......................................\left( 3 \right) \\\ & \left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}..........................\left( 4 \right) \\\ & \tan 3\theta =\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta }........................\left( 5 \right) \\\ \end{aligned}
Now, we will be using the above formulas for solving this question.
Now, first, we will find the value of tan⁡(1200−θ)\tan \left( {{120}^{0}}-\theta \right) in terms of tan⁡θ\tan \theta with the help of formula in the equation (2). Then,
tan⁡(1200−θ)=tan⁡1200−tan⁡θ1+tan⁡1200tan⁡θ\tan \left( {{120}^{0}}-\theta \right)=\dfrac{\tan {{120}^{0}}-\tan \theta }{1+\tan {{120}^{0}}\tan \theta }
Now, put tan⁡1200=−3\tan {{120}^{0}}=-\sqrt{3} in the above equation from the equation (3). Then,
tan⁡(1200−θ)=tan⁡1200−tan⁡θ1+tan⁡1200tan⁡θ ⇒tan⁡(1200−θ)=−3−tan⁡θ1−3tan⁡θ ⇒tan⁡(1200−θ)=−(3+tan⁡θ)(1−3tan⁡θ) ⇒tan⁡(1200−θ)=(3+tan⁡θ)(3tan⁡θ−1)......................(6) \begin{aligned} & \tan \left( {{120}^{0}}-\theta \right)=\dfrac{\tan {{120}^{0}}-\tan \theta }{1+\tan {{120}^{0}}\tan \theta } \\\ & \Rightarrow \tan \left( {{120}^{0}}-\theta \right)=\dfrac{-\sqrt{3}-\tan \theta }{1-\sqrt{3}\tan \theta } \\\ & \Rightarrow \tan \left( {{120}^{0}}-\theta \right)=-\dfrac{\left( \sqrt{3}+\tan \theta \right)}{\left( 1-\sqrt{3}\tan \theta \right)} \\\ & \Rightarrow \tan \left( {{120}^{0}}-\theta \right)=\dfrac{\left( \sqrt{3}+\tan \theta \right)}{\left( \sqrt{3}\tan \theta -1 \right)}......................\left( 6 \right) \\\ \end{aligned}
Now, first, we will find the value of tan⁡(1200+θ)\tan \left( {{120}^{0}}+\theta \right) in terms of tan⁡θ\tan \theta with the help of formula in the equation (1). Then,
tan⁡(1200+θ)=tan⁡1200+tan⁡θ1−tan⁡1200tan⁡θ\tan \left( {{120}^{0}}+\theta \right)=\dfrac{\tan {{120}^{0}}+\tan \theta }{1-\tan {{120}^{0}}\tan \theta }
Now, put tan⁡1200=−3\tan {{120}^{0}}=-\sqrt{3} in the above equation from the equation (3). Then,
tan⁡(1200+θ)=tan⁡1200+tan⁡θ1−tan⁡1200tan⁡θ ⇒tan⁡(1200+θ)=−3+tan⁡θ1−(−3)tan⁡θ ⇒tan⁡(1200+θ)=tan⁡θ−31+3tan⁡θ.................(7) \begin{aligned} & \tan \left( {{120}^{0}}+\theta \right)=\dfrac{\tan {{120}^{0}}+\tan \theta }{1-\tan {{120}^{0}}\tan \theta } \\\ & \Rightarrow \tan \left( {{120}^{0}}+\theta \right)=\dfrac{-\sqrt{3}+\tan \theta }{1-\left( -\sqrt{3} \right)\tan \theta } \\\ & \Rightarrow \tan \left( {{120}^{0}}+\theta \right)=\dfrac{\tan \theta -\sqrt{3}}{1+\sqrt{3}\tan \theta }.................\left( 7 \right) \\\ \end{aligned}
Now, we will solve for tan⁡θ⋅tan⁡(1200−θ)⋅tan⁡(1200+θ)\tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right) by substituting tan⁡(1200−θ)=(3+tan⁡θ)(3tan⁡θ−1)\tan \left( {{120}^{0}}-\theta \right)=\dfrac{\left( \sqrt{3}+\tan \theta \right)}{\left( \sqrt{3}\tan \theta -1 \right)} from equation (6) and tan⁡(1200+θ)=tan⁡θ−31+3tan⁡θ\tan \left( {{120}^{0}}+\theta \right)=\dfrac{\tan \theta -\sqrt{3}}{1+\sqrt{3}\tan \theta } from equation (7). Then,
tan⁡θ⋅tan⁡(1200−θ)⋅tan⁡(1200+θ) ⇒tan⁡θ⋅tan⁡(1200−θ)⋅tan⁡(1200+θ)=tan⁡θ×(3+tan⁡θ)(3tan⁡θ−1)×(tan⁡θ−3)(1+3tan⁡θ) ⇒tan⁡θ⋅tan⁡(1200−θ)⋅tan⁡(1200+θ)=tan⁡θ×(tan⁡θ+3)×(tan⁡θ−3)(3tan⁡θ+1)×(3tan⁡θ−1) \begin{aligned} & \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right) \\\ & \Rightarrow \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\tan \theta \times \dfrac{\left( \sqrt{3}+\tan \theta \right)}{\left( \sqrt{3}\tan \theta -1 \right)}\times \dfrac{\left( \tan \theta -\sqrt{3} \right)}{\left( 1+\sqrt{3}\tan \theta \right)} \\\ & \Rightarrow \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\tan \theta \times \dfrac{\left( \tan \theta +\sqrt{3} \right)\times \left( \tan \theta -\sqrt{3} \right)}{\left( \sqrt{3}\tan \theta +1 \right)\times \left( \sqrt{3}\tan \theta -1 \right)} \\\ \end{aligned}
Now, use the formula (a+b)(a−b)=a2−b2\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} form equation (4) to write (tan⁡θ+3)×(tan⁡θ−3)=tan⁡2θ−3\left( \tan \theta +\sqrt{3} \right)\times \left( \tan \theta -\sqrt{3} \right)={{\tan }^{2}}\theta -3 and (3tan⁡θ+1)×(3tan⁡θ−1)=3tan⁡2θ−1\left( \sqrt{3}\tan \theta +1 \right)\times \left( \sqrt{3}\tan \theta -1 \right)=3{{\tan }^{2}}\theta -1 in the above equation. Then,
tan⁡θ⋅tan⁡(1200−θ)⋅tan⁡(1200+θ)=tan⁡θ×(tan⁡θ+3)×(tan⁡θ−3)(3tan⁡θ+1)×(3tan⁡θ−1) ⇒tan⁡θ⋅tan⁡(1200−θ)⋅tan⁡(1200+θ)=tan⁡θ×(tan⁡2θ−3)(3tan⁡2θ−1) ⇒tan⁡θ⋅tan⁡(1200−θ)⋅tan⁡(1200+θ)=(tan⁡3θ−3tan⁡θ)(3tan⁡2θ−1) ⇒tan⁡θ⋅tan⁡(1200−θ)⋅tan⁡(1200+θ)=−(3tan⁡θ−tan⁡3θ)(3tan⁡2θ−1) ⇒tan⁡θ⋅tan⁡(1200−θ)⋅tan⁡(1200+θ)=(3tan⁡θ−tan⁡3θ)(1−3tan⁡2θ) \begin{aligned} & \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\tan \theta \times \dfrac{\left( \tan \theta +\sqrt{3} \right)\times \left( \tan \theta -\sqrt{3} \right)}{\left( \sqrt{3}\tan \theta +1 \right)\times \left( \sqrt{3}\tan \theta -1 \right)} \\\ & \Rightarrow \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\tan \theta \times \dfrac{\left( {{\tan }^{2}}\theta -3 \right)}{\left( 3{{\tan }^{2}}\theta -1 \right)} \\\ & \Rightarrow \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\dfrac{\left( {{\tan }^{3}}\theta -3\tan \theta \right)}{\left( 3{{\tan }^{2}}\theta -1 \right)} \\\ & \Rightarrow \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=-\dfrac{\left( 3\tan \theta -{{\tan }^{3}}\theta \right)}{\left( 3{{\tan }^{2}}\theta -1 \right)} \\\ & \Rightarrow \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\dfrac{\left( 3\tan \theta -{{\tan }^{3}}\theta \right)}{\left( 1-3{{\tan }^{2}}\theta \right)} \\\ \end{aligned}
Now, from equation (5) we know that, tan⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ\tan 3\theta =\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta } . Then,
tan⁡θ⋅tan⁡(1200−θ)⋅tan⁡(1200+θ)=(3tan⁡θ−tan⁡3θ)(1−3tan⁡2θ) ⇒tan⁡θ⋅tan⁡(1200−θ)⋅tan⁡(1200+θ)=tan⁡3θ \begin{aligned} & \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\dfrac{\left( 3\tan \theta -{{\tan }^{3}}\theta \right)}{\left( 1-3{{\tan }^{2}}\theta \right)} \\\ & \Rightarrow \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\tan 3\theta \\\ \end{aligned}
Now, as it is given that tan⁡θ⋅tan⁡(1200−θ)⋅tan⁡(1200+θ)=13\tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\dfrac{1}{\sqrt{3}} . Then,
tan⁡θ⋅tan⁡(1200−θ)⋅tan⁡(1200+θ)=tan⁡3θ ⇒tan⁡3θ=13 \begin{aligned} & \tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\tan 3\theta \\\ & \Rightarrow \tan 3\theta =\dfrac{1}{\sqrt{3}} \\\ \end{aligned}
Now, we know that tan⁡π6=13\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}} . Then,
tan⁡3θ=13 ⇒tan⁡3θ=tan⁡π6..........................(8) \begin{aligned} & \tan 3\theta =\dfrac{1}{\sqrt{3}} \\\ & \Rightarrow \tan 3\theta =\tan \dfrac{\pi }{6}..........................\left( 8 \right) \\\ \end{aligned}
Now, before we proceed we should know one important result which we will use here.
If tan⁡x=tan⁡y\tan x=\tan y , then the general solution for xx in terms of y can be written as,
x=nπ+y....................(9)x=n\pi +y....................\left( 9 \right) , where nn is any integer.
From equation (8) we know that, tan⁡3θ=tan⁡π6\tan 3\theta =\tan \dfrac{\pi }{6} . So, we can use the formula from the equation (9). Then,
tan⁡3θ=tan⁡π6 ⇒3θ=nπ+π6,n∈z ⇒θ=nπ3+π18,n∈z \begin{aligned} & \tan 3\theta =\tan \dfrac{\pi }{6} \\\ & \Rightarrow 3\theta =n\pi +\dfrac{\pi }{6},n\in z \\\ & \Rightarrow \theta =\dfrac{n\pi }{3}+\dfrac{\pi }{18},n\in z \\\ \end{aligned}
Now, from the above result, we conclude that, if tan⁡θ⋅tan⁡(1200−θ)⋅tan⁡(1200+θ)=13\tan \theta \cdot \tan \left( {{120}^{0}}-\theta \right)\cdot \tan \left( {{120}^{0}}+\theta \right)=\dfrac{1}{\sqrt{3}} , then
θ=nπ3+π18,n∈z\theta =\dfrac{n\pi }{3}+\dfrac{\pi }{18},n\in z .
Hence, (c) will be the correct option.

Note: Here, the student should first understand what is asked in the problem and then proceed in the right direction to get the correct answer. And we should remember the results like tan⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ\tan 3\theta =\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta } and solution of the equation tan⁡x=tan⁡y\tan x=\tan y correctly and apply them without any mistake while solving. Moreover, we should avoid making calculation mistakes and after writing the final answer for such questions we should select the correct option only.