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Question: Solve and find the value of the following: \(\tan 3\theta =-1\) A)\(\theta =\dfrac{n\pi }{3}-\dfra...

Solve and find the value of the following: tan⁡3θ=−1\tan 3\theta =-1
A)θ=nπ3−π6,n∈Z\theta =\dfrac{n\pi }{3}-\dfrac{\pi }{6},n\in Z
B)θ=nπ3−π3,n∈Z\theta =\dfrac{n\pi }{3}-\dfrac{\pi }{3},n\in Z
C)θ=nπ3−π4,n∈Z\theta =\dfrac{n\pi }{3}-\dfrac{\pi }{4},n\in Z
D)θ=nπ3−π12,n∈Z\theta =\dfrac{n\pi }{3}-\dfrac{\pi }{12},n\in Z

Explanation

Solution

The question can be solved using the general trigonometric equations and using tan⁡π4=−1\tan \dfrac{\pi }{4}=-1.
And tan⁡(π−θ)=−tan⁡θ\tan \left( \pi -\theta \right)=-\tan \theta .
Simplify the expression to obtain the value of θ\theta . The tan⁡\tan function is negative in the second and fourth quadrant therefore consider this while solving the question.

Complete step by step solution:
Given,
tan⁡3θ=−1\tan 3\theta =-1
We can write tan⁡π4=−1\tan \dfrac{\pi }{4}=-1 in the above equation, we get:
tan⁡3θ=−tan⁡π4\tan 3\theta =-\tan \dfrac{\pi }{4}
Now we can use the property tan⁡(π−θ)=−tan⁡θ\tan \left( \pi -\theta \right)=-\tan \theta and replace the term −tan⁡π4-\tan \dfrac{\pi }{4} in the expression.
⇒tan⁡3θ=tan⁡(π−π4) ⇒tan⁡3θ=tan⁡(3π4) \begin{aligned} & \Rightarrow \tan 3\theta =\tan \left( \pi -\dfrac{\pi }{4} \right) \\\ & \Rightarrow \tan 3\theta =\tan \left( \dfrac{3\pi }{4} \right) \\\ \end{aligned}
Now since there are only tan⁡\tan terms in left-hand side and right-hand side therefore we can write the general solution of the equation as follows:
3θ=nπ−3π4 θ=nπ3−3π4×3 ∴θ=nπ3−π4 \begin{aligned} & 3\theta =n\pi -\dfrac{3\pi }{4} \\\ & \theta =\dfrac{n\pi }{3}-\dfrac{3\pi }{4\times 3} \\\ & \therefore \theta =\dfrac{n\pi }{3}-\dfrac{\pi }{4} \\\ \end{aligned}

So, the correct answer is “Option C”.

Note: There is one more method to solve the above question by general analysis:
The value of the function tan⁡\tan is positive in the first and third quadrant.
For θ=45∘\theta ={{45}^{\circ }} the value is 1.
For tan⁡\tan to be -1 the angle would be (π−π4),(2π−π4)\left( \pi -\dfrac{\pi }{4} \right),\left( 2\pi -\dfrac{\pi }{4} \right)
The equation that involves trigonometric equations are called trigonometric equations.
The function tan⁡x\tan x repeats itself after an interval of π\pi .
For a general equation tan⁡x=0\tan x=0 the possible solution of equation is x=nπx=n\pi .
In the above question we have tan⁡3θ=−1\tan 3\theta =-1 therefore we apply the necessary properties and most importantly we must remember that the value of the function is positive in the first and third quadrant.
There is one more method to solve the above question by general analysis:
The value of the function tan⁡\tan is positive in the first and third quadrant.
For θ=45∘\theta ={{45}^{\circ }} the value is 1.