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Question: The general solution of tan\(3\)θ tan θ = \(1\) is given by (n \(\in \) I). θ = \((2n+1)\frac{\pi ...

The general solution of tan33θ tan θ = 11 is given by (n ∈\in I).
θ = (2n+1)π2(2n+1)\frac{\pi }{2}
θ = (2n+1)π4(2n+1)\frac{\pi }{4}
θ = nπ±π8,nπ±3π8n\pi \pm \frac{\pi }{8},n\pi \pm \frac{3\pi }{8}
θ = (2n+1)π8(2n+1)\frac{\pi }{8}

Explanation

Solution

The principal and general solutions of given trigonometric ratio refers to general form of value of given trigonometric ratio.
The value of trigonometric ratios is always calculated in a right-angled triangle.

Complete step by step answer:
There are four quadrants in which the plane is divided. All six trigonometric ratios are positive in first quadrant, sinθ and cosecθ are positive in second quadrant while other trigonometric ratios are negative, tanθ and cotθ are positive in third quadrant while other trigonometric ratios are negative and cosθ and secθ are positive in fourth quadrant while other trigonometric ratios are negative.
The trigonometric ratio tanθ = sin⁡θcos⁡θ\frac{\sin \theta }{\cos \theta } .
Putting the value of tanθ and tan33θ in terms of sin and cos in the equation gives:

& \tan \theta \tan 3\theta =1 \\\ & \frac{\sin \theta \sin 3\theta }{\cos \theta \cos 3\theta }=1 \\\ & \sin \theta \sin 3\theta =\cos \theta \cos 3\theta \\\ & co\operatorname{s}\theta co\operatorname{s}3\theta -\sin \theta \sin 3\theta =0 \end{aligned}$$ As cosA cosB – sinA sinB = cos (A + B, the value is substituted in equation derived above as: $$\begin{aligned} & co\operatorname{s}\theta co\operatorname{s}3\theta -\sin \theta \sin 3\theta =0 \\\ & \cos (\theta +3\theta )=0 \\\ & \cos 4\theta =0 \end{aligned}$$ The general solution of equation cosθ = cosα is given by: θ = $2n\pi \pm \alpha $ This implies that general solution for equation $\cos 4\theta =0$ is given as: $\begin{aligned} & \theta =\left(\dfrac{ 2n\pi}{4} \pm \frac{\pi }{4\times 2} \right) \\\ & =\left( \dfrac{ 2n\pi}{4} \pm \frac{\pi }{8} \right) \end{aligned}$ As angle $0{}^\circ $ lies in first quadrant, the general solution of equation, $$\tan \theta \tan 3\theta =1$$ is given as: $\begin{aligned} & \theta =\left(\dfrac{ 2n\pi}{4} +\frac{\pi }{8} \right) \\\ \end{aligned}$ **So, the correct answer is “Option D”.** **Note:** There are $6$ main trigonometric ratios in all. There are three pairs of trigonometric ratios – (sinθ, cosecθ), (cosθ, secθ) and (tanθ, cotθ). The trigonometric ratio is written as sinθ which means the ratio of perpendicular to Hypotenuse but it does not mean product of sin and θ. This indicates that option D is correct.